第12类RD Sharma解决方案-第19章不定积分–练习19.7

积分以下积分：

问题1.∫sin4xcos7x dx

解决方案：

Let I= $\int \sin4x\cos7x\,dx$

We know,

$2\sin A \cos B= \sin(A+B)+\sin(A-B)$

Applying this formula to the given question we get

I=$\int \frac 1 2(\sin(4x+7x)+\sin(4x-7x))\,dx$

= $\int \frac 1 2(\sin11x+\sin(-3x)\,dx$  

=$\int\frac 1 2 (\sin 11x -\sin3x)\,dx$  $[\because \sin(-\theta)=-\sin\theta]$

=$\frac 1 2(\int\sin 11x -\int\sin3x)\,dx$

We know,

$\int \sin ax\,dx=-\frac 1 a\cos ax+C$

Applying this formula to the given question we get

I= $\frac 1 2(-\frac {1} {11} \cos 11x +\frac 1 3 \cos3x)$

$\therefore$ I= $-\frac {1} {22} \cos 11x +\frac 1 6 \cos3x+C$

问题2。∫cos3x cos4x dx

解决方案：

Let I= $\int \cos3x\cos4x\,dx$

Multiplying and dividing the equation by 2,we get

I=$\frac 1 2\int 2\cos3x\cos4x\,dx$

We know,

$2\cos A \cos B= \cos(A+B)+\cos(A-B)$

Applying this formula to the given question we get

I=$\frac 1 2\int( \cos(3x+4x)+\cos(3x-4x))\,dx$

=$\frac 1 2\int( \cos 7x+\cos x)\,dx$ $[\because \cos(-\theta)=\cos\theta]$

We know,

$\int \cos x\,dx=\sin x+C$  and $\int \cos ax\,dx=\frac 1 a\sin ax+C$

Applying these formulas to the given question we get

I=$\frac 1 2 (\frac {\sin 7x} 7+\sin x) + C$

$\therefore$ I=$(\frac {\sin 7x} {14}+\frac {\sin x} 2) + C$

问题3.∫ COS MX COS NX DX，米≠N

解决方案：

Let I= $\int \cos mx \cos nx\,dx , m \ne n$

Multiplying and dividing the equation by 2,we get

I=$\frac 1 2\int 2\cos mx\cos nx\,dx$

We know,

$2\cos A \cos B= \cos(A+B)+\cos(A-B)$

Applying this formula to the given question we get

I=$\frac 1 2\int( \cos(m+n)x+\cos(m-n)x)\,dx$

We know,

$\int \cos ax\,dx=\frac 1 a\sin ax+C$

Applying these formulas to the given question we get

$\therefore$ I=$\frac 1 2 (\frac {\sin (m+n)x} {m+n}+\frac {\sin (m-n)x} {m-n}) + C$

问题4.∫sin mx cos nx dx ， m≠n

解决方案：

Let I=$\int \sin mx \cos nx\,dx , m \ne n$

Multiplying and dividing the equation by 2,we get

I=$\frac 1 2\int 2\sin mx\cos nx\,dx$

We know,

$2\sin A \cos B= \sin(A+B)+\sin(A-B)$

Applying this formula to the given question we get

I=$\frac 1 2\int( \sin(m+n)x+\sin(m-n)x)\,dx$

We know,

$\int \sin ax\,dx=-\frac 1 a\cos ax+C$

Applying these formulas to the given question we get

$\therefore$ I= $\frac 1 2 (-\frac {\cos (m+n)x} {m+n}-\frac {\cos (m-n)x} {m-n}) + C$