第 12 类 RD Sharma 解 – 第 20 章定积分 – 练习 20.2 |设置 1

计算以下定积分：

问题 1。 $\int_{2}^{4}\frac{x}{x^2+1}dx$

解决方案：

We have,

I = $\int_{2}^{4}\frac{x}{x^2+1}dx$

I = $\frac{1}{2}\int_{2}^{4}\frac{2x}{x^2+1}dx$

I = $\left[\frac{1}{2}\log(1+x^2)\right]_{2}^{4}$

I = $\frac{1}{2}\log(1+4^2)-\frac{1}{2}\log(1+2^2)$

I = $\frac{1}{2}\log(1+16)-\frac{1}{2}\log(1+4)$

I = $\frac{1}{2}\log17-\frac{1}{2}\log5$

I = $\frac{1}{2}(\log17-log5)$

I = $\frac{1}{2}\log\frac{17}{5}$

Therefore, the value of $\int_{2}^{4}\frac{x}{x^2+1}dx$ is $\frac{1}{2}\log\frac{17}{5}$.

问题2。 $\int_{1}^{2}\frac{1}{x(1+logx)^2}dx$

解决方案：

We have,

I = $\int_{1}^{2}\frac{1}{x(1+logx)^2}dx$

Let 1 + log x = t, so we have,

=> (1/x) dx = 2t dt

Now, the lower limit is, x = 1

=> t = 1 + log x

=> t = 1 + log 1

=> t = 1 + 0

=> t = 1

Also, the upper limit is, x = 2

=> t = 1 + log x

=> t = 1 + log 2

So, the equation becomes,

I = $\int_{1}^{1+\log2}\frac{1}{t^2}dt$

I = $\left[\frac{-1}{t}\right]_1^{1+\log2}$

I = $\left[\frac{-1}{1+\log2}+1\right]$

I = $\frac{-1+1+\log2}{1+\log2}$

I = $\frac{\log2}{1+\log2}$

I = $\frac{\log2}{\log e+\log2}$

I = $\frac{\log2}{\log2e}$

Therefore, the value of $\int_{1}^{2}\frac{1}{x(1+logx)^2}dx$ is $\frac{\log2}{\log2e}$.

问题 3。 $\int_{1}^{2}\frac{3x}{9x^2-1}dx$

解决方案：

We have,

I = $\int_{1}^{2}\frac{3x}{9x^2-1}dx$

Let 9x² – 1 = t, so we have,

=> 18x dx = dt

=> 3x dx = dt/6

Now, the lower limit is, x = 1

=> t = 9x² – 1

=> t = 9 (1)² – 1

=> t = 9 – 1

=> t = 8

Also, the upper limit is, x = 2

=> t = 9x² – 1

=> t = 9 (2)² – 1

=> t = 36 – 1

=> t = 35

So, the equation becomes,

I = $\int_{8}^{35}\frac{1}{6t}dt$

I = $\left[\frac{1}{6}\log t\right]_{8}^{35}$

I = $\frac{1}{6}(\log 35-\log 8)$

I = $\frac{1}{6}\log \frac{35}{8}$

Therefore, the value of $\int_{1}^{2}\frac{3x}{9x^2-1}dx$ is $\frac{1}{6}\log \frac{35}{8}$.

问题 4。 $\int_{0}^{\frac{\pi}{2}}\frac{1}{5cosx+3sinx}dx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{5cosx+3sinx}dx$

On putting sin x = $\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}$ and cos x = $\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}$, we get

I = $\int_{0}^{\frac{\pi}{2}}\frac{sec^2\frac{x}{2}}{5(1-tan^2\frac{x}{2})+6tan\frac{x}{2}}dx$

I = $\int_{0}^{\frac{\pi}{2}}\frac{sec^2\frac{x}{2}}{5-5tan^2\frac{x}{2}+6tan\frac{x}{2}}dx$

Let tan x/2 = t. So, we have

=> $\frac{1}{2}sec^2\frac{x}{2}$ = dt

Now, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x/2

=> t = tan π/4

=> t = 1

So, the equation becomes,

I = $\int_{0}^{1}\frac{2}{5-5t^2+6t}dt$

I = $\frac{2}{5}\int_{0}^{1}\frac{1}{1-t^2+\frac{6}{5}t}dt$

I = $\frac{2}{5}\int_{0}^{1}\frac{1}{\frac{34}{25}-(t-\frac{3}{5})^2}dt$

I = $\left[\frac{2}{5}.\frac{1}{2}\sqrt{\frac{25}{34}}\log\left(\frac{\sqrt{\frac{34}{25}}+t-\frac{3}{5}}{\sqrt{\frac{34}{25}}-t+\frac{3}{5}}\right)\right]^1_0$

I = $\left[\frac{2}{5}.\frac{1}{2}\frac{5}{\sqrt{34}}\log\left(\frac{\sqrt{\frac{34}{25}}+t-\frac{3}{5}}{\sqrt{\frac{34}{25}}-t+\frac{3}{5}}\right)\right]^1_0$

I = $\frac{1}{\sqrt{34}}\left[\log\left(\frac{\sqrt{34}+2}{\sqrt{34}-2}\right)-\log\left(\frac{\sqrt{34}-3}{\sqrt{34}+3}\right)\right]$

I = $\frac{1}{\sqrt{34}}\log\left(\frac{(\sqrt{34}+2)(\sqrt{34}+3)}{(\sqrt{34}-2)(\sqrt{34}-3)}\right)$

I = $\frac{1}{\sqrt{34}}\log\left(\frac{40+5\sqrt{34}}{40-5\sqrt{34}}\right)$

I = $\frac{1}{\sqrt{34}}\log\left(\frac{8+\sqrt{34}}{8-\sqrt{34}}\right)$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{1}{5cosx+3sinx}dx$ is $\frac{1}{\sqrt{34}}\log\left(\frac{8+\sqrt{34}}{8-\sqrt{34}}\right)$.

问题 5。 $\int_{0}^{a}\frac{x}{\sqrt{a^2+x^2}}dx$

解决方案：

We have,

I = $\int_{0}^{a}\frac{x}{\sqrt{a^2+x^2}}dx$

Let a² + x² = t². So, we have

=> 2x dx = 2t dt

=> x dx = t dt

Now, the lower limit is, x = 0

=> t² = a² + x²

=> t² = a² + 0²

=> t² = a²

=> t = a

Also, the upper limit is, x = a

=> t² = a² + x²

=> t² = a² + a²

=> t² = 2a²

=> t = √2 a

So, the equation becomes,

I = $\int_{a}^{\sqrt{2}a}\frac{t}{t}dt$

I = $\int_{a}^{\sqrt{2}a}dt$

I = $\left[t\right]_{a}^{\sqrt{2}a}$

I = √2a – a

I = a (√2 – 1)

Therefore, the value of $\int_{0}^{a}\frac{x}{\sqrt{a^2+x^2}}dx$ is a (√2 – 1).

问题 6。 $\int_{0}^{1}\frac{e^x}{1+e^{2x}}dx$

解决方案：

We have,

I = $\int_{0}^{1}\frac{e^x}{1+e^{2x}}dx$

Let e^x = t. So, we have

=> e^x dx = dt

Now, the lower limit is, x = 0

=> t = e^x

=> t = e⁰

=> t = 1

Also, the upper limit is, x = a

=> t = e^x

=> t = e¹

=> t = e

So, the equation becomes,

I = $\int_{1}^{e}\frac{1}{1+t^2}dt$

I = $\left[tan^{-1}t\right]_{1}^{e}$

I = $tan^{-1}e-tan^{-1}1$

I = $tan^{-1}e-\frac{\pi}{4}$

Therefore, the value of $\int_{0}^{1}\frac{e^x}{1+e^{2x}}dx$ is $tan^{-1}e-\frac{\pi}{4}$.

问题 7。 $\int_{0}^{1}xe^{x^2}dx$

解决方案：

We have,

I = $\int_{0}^{1}xe^{x^2}dx$

Let x² = t. So, we have

=> 2x dx = dt

Now, the lower limit is, x = 0

=> t = x²

=> t = 0²

=> t = 0

Also, the upper limit is, x = 1

=> t = x²

=> t = 1²

=> t = 1

So, the equation becomes,

I = $\int_{0}^{1}\frac{e^t}{2}dt$

I = $\frac{1}{2}\int_{0}^{1}e^tdt$

I = $\frac{1}{2}\left[e^t\right]_{0}^{1}$

I = $\frac{1}{2}\left[e^1-e^0\right]$

I = $\frac{1}{2}(e-1)$

Therefore, the value of $\int_{0}^{1}xe^{x^2}dx$ is $\frac{1}{2}(e-1)$.

问题 8。 $\int_{1}^{3}\frac{cos(logx)}{x}dx$

解决方案：

We have,

I = $\int_{1}^{3}\frac{cos(logx)}{x}dx$

Let log x = t. So, we have

=> (1/x) dx = dt

Now, the lower limit is, x = 1

=> t = log x

=> t = log 1

=> t = 0

Also, the upper limit is, x = 3

=> t = log x

=> t = log 3

So, the equation becomes,

I = $\int_{0}^{\log3}costdt$

I = $\left[sint\right]_{0}^{\log3}$

I = sin (log 3) – sin 0

I = sin (log 3) – 0

I = sin (log 3)

Therefore, the value of $\int_{1}^{3}\frac{cos(logx)}{x}dx$ is sin (log 3).

问题 9。 $\int_{0}^{1}\frac{2x}{1+x^4}dx$

解决方案：

We have,

I = $\int_{0}^{1}\frac{2x}{1+x^4}dx$

Let x² = t. So, we have

=> 2x dx = dt

Now, the lower limit is, x = 0

=> t = x²

=> t = 0²

=> t = 0

Also, the upper limit is, x = 1

=> t = x²

=> t = 1²

=> t = 1

So, the equation becomes,

I = $\int_{0}^{1}\frac{1}{1+t^2}dt$

I = $\left[tan^{-1}t\right]_{0}^{1}$

I = $tan^{-1}1-tan^{-1}0$

I = $\frac{\pi}{4}-0$

I = $\frac{\pi}{4}$

Therefore, the value of $\int_{0}^{1}\frac{2x}{1+x^4}dx$ is $\frac{\pi}{4}$.

问题 10。 $\int_{0}^{a}\sqrt{a^2-x^2}dx$

解决方案：

We have,

I = $\int_{0}^{a}\sqrt{a^2-x^2}dx$

Let x = a sin t. So, we have

=> dx = a cos t dt

Now, the lower limit is, x = 0

=> a sin t = x

=> a sin t = 0

=> sin t = 0

=> t = 0

Also, the upper limit is, x = a

=> a sin t = a

=> a sin t = a

=> sin t = 1

=> t = π/2

So, the equation becomes,

I = $\int_{0}^{\frac{\pi}{2}}\sqrt{a^2-a^2sin^2t}(acost)dt$

I = $\int_{0}^{\frac{\pi}{2}}\sqrt{a^2(1-sin^2t)}(acost)dt$

I = $\int_{0}^{\frac{\pi}{2}}\sqrt{a^2cos^2t}(acost)dt$

I = $\int_{0}^{\frac{\pi}{2}}(acost)(acost)dt$

I = $\int_{0}^{\frac{\pi}{2}}a^2cos^2tdt$

I = $a^2\int_{0}^{\frac{\pi}{2}}cos^2tdt$

I = $\frac{a^2}{2}\int_{0}^{\frac{\pi}{2}}(1+cos2t)dt$

I = $\frac{a^2}{2}\left[t+\frac{sin2t}{2}\right]_{0}^{\frac{\pi}{2}}$

I = $\frac{a^2}{2}\left[\frac{\pi}{2}+0-0-0\right]$

I = $\frac{a^2}{2}\left[\frac{\pi}{2}\right]$

I = $\frac{\pi a^2}{4}$

Therefore, the value of $\int_{0}^{a}\sqrt{a^2-x^2}dx$ is $\frac{\pi a^2}{4}$.

问题 11。 $\int_{0}^{\frac{\pi}{2}}\sqrt{sinØ}cos^5ØdØ$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{2}}\sqrt{sinØ}cos^5ØdØ$

I = $\int_{0}^{\frac{\pi}{2}}\sqrt{sinØ}cos^4ØcosØdØ$

Let sin Ø = t. So, we have

=> cos Ø dØ = dt

Now, the lower limit is, Ø = 0

=> t = sin Ø

=> t = sin 0

=> t = 0

Also, the upper limit is, Ø = π/2

=> t = sin Ø

=> t = sin π/2

=> t = 1

So, the equation becomes,

I = $\int_{0}^{1}\sqrt{t}(1-t^2)^2dt$

I = $\int_{0}^{1}\sqrt{t}(1+t^4-2t^2)dt$

I = $\int_{0}^{1}(t^\frac{1}{2}+t^\frac{9}{2}-2t^\frac{5}{2})dt$

I = $\left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{t^{\frac{9}{2}+1}}{\frac{9}{2}+1}-\frac{2t^{\frac{5}{2}+1}}{\frac{5}{2}+1}\right]_{0}^{1}$

I = $\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{t^{\frac{11}{2}}}{\frac{11}{2}}-\frac{2t^{\frac{7}{2}}}{\frac{7}{2}}\right]_{0}^{1}$

I = $\left[\frac{2t^{\frac{3}{2}}}{3}+\frac{2t^{\frac{11}{2}}}{11}-\frac{4t^{\frac{7}{2}}}{7}\right]_{0}^{1}$

I = $\frac{2}{3}+\frac{2}{11}-\frac{4}{7}$

I = $\frac{154+42-132}{231}$

I = $\frac{64}{231}$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\sqrt{sinØ}cos^5ØdØ$ is $\frac{64}{231}$.

问题 12。 $\int_{0}^{\frac{\pi}{2}}\frac{cosx}{1+sinx}dx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{2}}\frac{cosx}{1+sin^2x}dx$

Let sin x = t. So, we have

=> cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin x

=> t = sin 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin x

=> t = sin π/2

=> t = 1

So, the equation becomes,

I = $\int_{0}^{1}\frac{1}{1+t^2}dt$

I = $\left[tan^{-1}t\right]_{0}^{1}$

I = $tan^{-1}1-tan^{-1}0$

I = $\frac{\pi}{4}-0$

I = $\frac{\pi}{4}$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{cosx}{1+sin^2x}dx$ is $\frac{\pi}{4}$.

问题 13。 $\int_{0}^{\frac{\pi}{2}}\frac{sin\theta}{\sqrt{1+cos\theta}}d\theta$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{2}}\frac{sin\theta}{\sqrt{1+cos\theta}}d\theta$

Let 1 + cos θ = t². So, we have

=> – sin θ dθ = 2t dt

=> sin θ dθ = –2t dt

Now, the lower limit is, θ = 0

=> t² = 1 + cos θ

=> t² = 1 + cos 0

=> t² = 1 + 1

=> t² = 2

=> t = √2

Also, the upper limit is, θ = π/2

=> t² = 1 + cos θ

=> t² = 1 + cos π/2

=> t² = 1 + 0

=> t² = 1

=> t = 1

So, the equation becomes,

I = $\int_{\sqrt{2}}^{1}\frac{-2t}{t}dt$

I = $\int_{\sqrt{2}}^{1}-2dt$

I = $-2\left[t\right]_{\sqrt{2}}^{1}$

I = $-2\left[1-\sqrt{2}\right]$

I = $2\left[\sqrt{2}-1\right]$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{sin\theta}{\sqrt{1+cos\theta}}d\theta$ is $2\left[\sqrt{2}-1\right]$.

问题 14。 $\int_{0}^{\frac{\pi}{3}}\frac{cosx}{3+4sinx}dx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{3}}\frac{cosx}{3+4sinx}dx$

Let 3 + 4 sin x = t. So, we have

=> 0 + 4 cos x dx = dt

=> 4 cos x dx = dt

=> cos x dx = dt/4

Now, the lower limit is, x = 0

=> t = 3 + 4 sin x

=> t = 3 + 4 sin 0

=> t = 3 + 0

=> t = 3

Also, the upper limit is, x = π/3

=> t = 3 + 4 sin x

=> t = 3 + 4 sin π/3

=> t = 3 + 4 (√3/2)

=> t = 3 + 2√3

So, the equation becomes,

I = $\int_{3}^{3+2\sqrt{3}}\frac{1}{4t}dt$

I = $\frac{1}{4}\left[\log t\right]_{3}^{3+2\sqrt{3}}$

I = $\frac{1}{4}\left[\log (3+2\sqrt{3})-\log 3)\right]$

I = $\frac{1}{4}\log\frac{3+2\sqrt{3}}{3}$

Therefore, the value of $\int_{0}^{\frac{\pi}{3}}\frac{cosx}{3+4sinx}dx$ is $\frac{1}{4}\log\frac{3+2\sqrt{3}}{3}$.

问题 15。 $\int_{0}^{1}\frac{\sqrt{tan^{-1}x}}{1+x^2}dx$

解决方案：

We have,

I = $\int_{0}^{1}\frac{\sqrt{tan^{-1}x}}{1+x^2}dx$

Let tan^–1 x = t. So, we have

=> (1/1+x²) dx = dt

Now, the lower limit is, x = 0

=> t = tan^–1 x

=> t = tan^–1 0

=> t = 0

Also, the upper limit is, x = 1

=> t = tan^–1 t

=> t = tan^–1 1

=> t = π/4

So, the equation becomes,

I = $\int_{0}^{\frac{\pi}{4}}t^{\frac{1}{2}}dt$

I = $\left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{\frac{\pi}{4}}$

I = $\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{\frac{\pi}{4}}$

I = $\frac{2}{3}\left[t^{\frac{3}{2}}\right]_{0}^{\frac{\pi}{4}}$

I = $\frac{2}{3}\left[(\frac{\pi}{4})^{\frac{3}{2}}-0\right]$

I = $\frac{2}{3}(\frac{\pi}{4})^{\frac{3}{2}}$

I = $\frac{2}{24}\pi^{\frac{3}{2}}$

I = $\frac{1}{12}\pi^{\frac{3}{2}}$

Therefore, the value of $\int_{0}^{1}\frac{\sqrt{tan^{-1}x}}{1+x^2}dx$ is $\frac{1}{12}\pi^{\frac{3}{2}}$.

问题 16。 $\int_{0}^{2}x\sqrt{x+2}dx$

解决方案：

We have,

I = $\int_{0}^{2}x\sqrt{x+2}dx$

Let x + 2 = t². So, we have

=> dx = 2t dt

Now, the lower limit is, x = 0

=> t² = x + 2

=> t² = 0 + 2

=> t² = 2

=> t = √2

Also, the upper limit is, x = 2

=> t² = x + 2

=> t² = 2 + 2

=> t² = 4

=> t = 2

So, the equation becomes,

I = $\int_{\sqrt{2}}^{2}(t^2-2)\sqrt{t^2}2tdt$

I = $2\int_{\sqrt{2}}^{2}(t^2-2)t^2dt$

I = $2\int_{\sqrt{2}}^{2}(t^4-2t^2)dt$

I = $2\left[\frac{t^5}{5}-\frac{2t^3}{3}\right]_{\sqrt{2}}^{2}$

I = $2\left[\frac{32}{5}-\frac{16}{3}-\frac{4\sqrt{2}}{5}+\frac{4\sqrt{2}}{3}\right]$

I = $2\left[\frac{16+8\sqrt{2}}{15}\right]$

I = $2\left[\frac{8\sqrt{2}(1+\sqrt{2})}{15}\right]$

I = $\frac{16\sqrt{2}(1+\sqrt{2})}{15}$

Therefore, the value of $\int_{0}^{2}x\sqrt{x+2}dx$ is $\frac{16\sqrt{2}(1+\sqrt{2})}{15}$.

问题 17。 $\int_{0}^{1}tan^{-1}\frac{2x}{1-x^2}dx$

解决方案：

We have,

I = $\int_{0}^{1}tan^{-1}\frac{2x}{1-x^2}dx$

Let x = tan t. So, we have

=> dx = sec² t dt

Now, the lower limit is, x = 0

=> tan t = x

=> tan x = 0

=> x = 0

Also, the upper limit is, x = 1

=> tan t = x

=> tan x = 1

=> x = π/4

So, the equation becomes,

I = $\int_{0}^{\frac{\pi}{4}}tan^{-1}(\frac{2tant}{1-tan^2t})sec^2tdt$

I = $\int_{0}^{\frac{\pi}{4}}tan^{-1}(tan2t)sec^2tdt$

I = $\int_{0}^{\frac{\pi}{4}}2tsec^2tdt$

I = $2\int_{0}^{\frac{\pi}{4}}tsec^2tdt$

On applying integration by parts method, we get

I = $2\left[t\int_0^\frac{\pi}{4}sec^2tdt-\int_0^\frac{\pi}{4}tantdt\right]$

I = $2\left[ttant-log(cost)\right]^{\frac{\pi}{4}}_0$

I = $2\left[\frac{\pi}{4}+log\frac{1}{\sqrt{2}}-0-0\right]$

I = $2\left[\frac{\pi}{4}+\frac{1}{2}log2\right]$

I = $\frac{\pi}{2}+log2$

Therefore, the value of $\int_{0}^{1}tan^{-1}\frac{2x}{1-x^2}dx$ is $\frac{\pi}{2}+log 2$.

问题 18。 $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{1+sin^4x}dx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{1+sin^4x}dx$

Let sin² x = t. So, we have

=> 2 sin x cos x = dt

=> sin x cos x = dt/2

Now, the lower limit is, x = 0

=> t = sin² x

=> t = sin² 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin² x

=> t = sin² π/2

=> t = 1

So, the equation becomes,

I = $\int_{0}^{1}\frac{1}{2(1+t^2)}dt$

I = $\frac{1}{2}\int_{0}^{1}\frac{1}{1+t^2}dt$

I = $\frac{1}{2}\left[tan^{-1}t\right]^1_0$

I = $\frac{1}{2}\left[tan^{-1}1-tan^{-1}0\right]$

I = $\frac{1}{2}\left[\frac{\pi}{4}-0\right]$

I = $\frac{1}{2}(\frac{\pi}{4})$

I = $\frac{\pi}{8}$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{1+sin^4x}dx$ is $\frac{\pi}{8}$.

问题 19。 $\int_{0}^{\frac{\pi}{2}}\frac{1}{acosx+bsinx}dx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{acosx+bsinx}dx$

On putting cos x = $\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{1-tan^2\frac{x}{2}}{sec^2\frac{x}{2}}$ and sin x = $\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{2tan\frac{x}{2}}{sec^2\frac{x}{2}}$, we get,

I = $\int_{0}^{\frac{\pi}{2}}\frac{sec^2\frac{x}{2}}{a(1-tan^2\frac{x}{2})+2btan\frac{x}{2}}dx$

Let tan x/2 = t. So, we have

=> 1/2 sec² x/2 dx = dt

=> sec² x/2 dx = 2 dt

Now, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x/2

=> t = tan π/4

=> t = 1

So, the equation becomes,

I = $2\int_{0}^{1}\frac{1}{a(1-t^2)+2bt}dt$

I = $2\int_{0}^{1}\frac{1}{a-at^2+2bt}dt$

I = $2\int_{0}^{1}\frac{1}{-a(t^2-\frac{2b}{a}t-1)}dt$

I = $\frac{2}{a}\int_{0}^{1}\frac{1}{-\left[(t-\frac{b}{a})^2-1-\frac{b^2}{a^2}\right]}dt$

I = $\frac{2}{a}\int_{0}^{1}\frac{1}{(\frac{b^2}{a^2}+1)-(t-\frac{b}{a})^2}dt$

I = $\left[\frac{2}{a}\left(\frac{1}{2\sqrt{\frac{b^2+a^2}{a^2}}}\right)\log\left(\frac{\sqrt{\frac{b^2+a^2}{a^2}}+(t-\frac{b}{a})}{\sqrt{\frac{b^2+a^2}{a^2}}-(t-\frac{b}{a})}\right)\right]_{0}^{1}$

I = $\frac{1}{\sqrt{b^2+a^2}}\log\left(\frac{a+b+\sqrt{a^2+b^2}}{a+b-\sqrt{a^2+b^2}}\right)$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{1}{acosx+bsinx}dx$ is $\frac{1}{\sqrt{b^2+a^2}}\log\left(\frac{a+b+\sqrt{a^2+b^2}}{a+b-\sqrt{a^2+b^2}}\right)$.

问题 20。 $\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4sinx}dx$

解决方案：

We have,

I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4sinx}dx$

On putting sin x = $\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}$, we get

I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4\left(\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}\right)}dx$

I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{5+\left(\frac{8tan\frac{x}{2}}{1+tan^2\frac{x}{2}}\right)}dx$

I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{\frac{5(1+tan^2\frac{x}{2})+8tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}dx$

I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{\frac{5+5tan^2\frac{x}{2}+8tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}dx$

I = $\int_{0}^{\frac{\pi}{2}}\frac{1}{\frac{5+5tan^2\frac{x}{2}+8tan\frac{x}{2}}{sec^2\frac{x}{2}}}dx$

I = $\int_{0}^{\frac{\pi}{2}}\frac{sec^2\frac{x}{2}}{5+5tan^2\frac{x}{2}+8tan\frac{x}{2}}dx$

Let tan x/2 = t. So, we have

=> 1/2 sec² x/2 dx = dt

=> sec² x/2 dx = 2 dt

Now, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x/2

=> t = tan π/4

=> t = 1

So, the equation becomes,

I = $\int_{0}^{1}\frac{1}{5+5t^2+8t}dt$

I = $\frac{2}{5}\int_{0}^{1}\frac{1}{1+t^2+\frac{8}{5}t}dt$

I = $\frac{2}{5}\int_{0}^{1}\frac{1}{1-\frac{16}{25}+\frac{16}{25}+t^2+\frac{8}{5}t}dt$

I = $\frac{2}{5}\int_{0}^{1}\frac{1}{\frac{9}{25}+\frac{16}{25}+t^2+\frac{8}{5}t}dt$

I = $\frac{2}{5}\int_{0}^{1}\frac{1}{(\frac{3}{5})^2+(t+\frac{4}{5})^2}dt$

I = $\frac{2}{5}\left[\frac{5}{3}tan^{-1}\frac{5}{3}(t+\frac{4}{5})\right]_{0}^{1}$

I = $\frac{2}{3}\left[tan^{-1}(\frac{5t}{3}+\frac{4}{3})\right]_{0}^{1}$

I = $\frac{2}{3}\left[tan^{-1}(\frac{5}{3}+\frac{4}{3})-tan^{-1}(0+\frac{4}{3})\right]$

I = $\frac{2}{3}\left[tan^{-1}(\frac{9}{3})-tan^{-1}(\frac{4}{3})\right]$

I = $\frac{2}{3}\left[tan^{-1}(3)-tan^{-1}(\frac{4}{3})\right]$

I = $\frac{2}{3}\left[tan^{-1}(\frac{3-\frac{4}{3}}{1+3(\frac{4}{3})})\right]$

I = $\frac{2}{3}\left[tan^{-1}(\frac{\frac{5}{3}}{1+4})\right]$

I = $\frac{2}{3}tan^{-1}(\frac{\frac{5}{3}}{5})$

I = $\frac{2}{3}tan^{-1}(\frac{1}{3})$

Therefore, the value of $\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4sinx}dx$ is $\frac{2}{3}tan^{-1}(\frac{1}{3})$.