第12类RD Sharma解–第19章不定积分–练习19.18 |套装1

问题1.评估∫X ^/√x4 + ⁴ DX

解决方案：

Let us assume I = ∫ x/ √x⁴+a⁴ dx

= ∫ x/ √(x²)²+(a²)² dx (i)

Put x² = t

2x dx = dt

x dx = dt/2

Put the above value in eq. (i)

= 1/2 ∫ dt/√t² +(a²)

Integrate the above eq. then, we get

= 1/2 log |t+ √t²+(a²)²| + c [since ∫ 1/√x²+a² dx = log|x +√x²+a²| + c]

= 1/2 log |x²+ √(x²)²+(a²)²| + c

Hence, I = 1/2 log |x²+ √x⁴+a⁴| + c

问题2.评估∫秒^{2×/√tan2×+} 4 DX

解决方案：

Let us assume I =∫ sec²x/ √tan²x+4 dx (i)

Put tan x = t

sec²x dx = dt

Put the above value in eq. (i)

= ∫ dt/ √t²+(2)²

Integrate the above eq. then, we get

= log|t +√t²+(2)²| + c [since ∫ 1/√x²+a² dx =log|x +√x²+a²| + c]

= log|tanx +√tan²x+(2)²| + c

Hence, I = log|tanx +√tan²x+4| + c

问题3.评估∫E ^X /√16-E ^2X DX

解决方案：

Let us assume I =∫ e^x/ √16-e^2x dx (i)

Put e^x = t

e^x dx = dt

Put the above value in eq. (i)

= ∫ dt/ √(4)²-(e)²

Integrate the above eq. then, we get

= sin^-1(t/4) + c [since ∫1/ √a² – x² dx = sin^-1(x/a) + c]

= sin^-1(e^x/4) + c

Hence, I = sin^-1(e^x/4) + c

问题4.评估∫cosx /√4+ sin ² x dx

解决方案：

Let us assume I =∫ cosx/ √4+sin²x dx (i)

Put sinx = t

cosx dx = dt

Put the above value in eq. (i)

= ∫ dt/ √(2)²+t²

Integrate the above eq. then, we get

= log|t +√(2)²+t²| + c [since ∫ 1/√x²+a² dx =log|x +√x²+a²| + c]

= log|sinx +√(2)²+sin²x| + c

Hence, I = log|sinx +√4+sin²x| + c

问题5.评估∫的SiNx ^/√4cos2 X-1 DX

解决方案：

Let us assume I =∫ sinx/ √4cos²x-1 dx (i)

Put 2cosx = t

-2sinx dx = dt

sinx dx = -dt/2

Put the above value in eq. (i)

= -1/2 ∫ dt/ √t²-(1)²

Integrate the above eq. then, we get

= -1/2 log|t +√t²-(1)²| + c [since ∫ 1/√x²-a² dx =log|x +√x²-a²| + c]

= -1/2 log|2cosx +√(2cosx)²-(1)²| + c

Hence, I = -1/2 log|2cosx +√4cos²x-1| + c

问题6.评估∫X ^/√4-×4 DX

解决方案：

Let us assume I =∫ x/ √4-x⁴ dx (i)

Put x² = t

2x dx = dt

x dx = dt/2

Put the above value in eq. (i)

=1/2 ∫ dt/ √(2)²-(t)²

Integrate the above eq. then, we get

= sin^-1(t/2) + c [ since ∫1/ √a² – x² dx = sin^-1(x/a) + c]

= sin^-1(x²/2) + c

Hence, I = sin^-1(x²/2) + c

问题7.评估∫1 /x√4-9(logx) ² dx

解决方案：

Let us assume I =∫ 1/ x√4-9(logx)² dx

=∫ 1/ x√4-(3logx)² dx (i)

Put 3logx = t

3/x dx = dt

1/x dx = dt/3

Put the above value in eq. (i)

=1/3 ∫ dt/ √4-t²

=1/3 ∫ dt/ √(2)²-t²

Integrate the above eq. then, we get

=1/3 sin^-1(t/2) + c [since ∫1/ √a² – x² dx = sin^-1(x/a) + c]

=1/3 sin^-1(3logx/2) + c

Hence, I =1/3 sin^-1(3logx/2) + c

问题8.评估∫sin8x /√9+ sin ⁴ 4x dx

解决方案：

Let us assume I =∫ sin8x/ √9+sin⁴4x dx (i)

Put sin²4x = t

2sin4xcos4x (4)dx = dt

4sin8x dx = dt

sin8x dx = dt/4

Put the above value in eq. (i)

= 1/4 ∫ dt/ √9+t²

= 1/4 ∫ dt/ √(3)²+t²

Integrate the above eq. then, we get

= 1/4 log|t +√(3)²+t²| + c [since ∫ 1/√a²+x² dx =log|x +√a²+x²| + c]

= 1/4 log|sin⁴4x +√(3)²+sin⁴4x| + c

Hence, I = 1/4 log|sin²4x +√9+sin⁴4x| + c

问题9.评估∫cos2x ^/√sin2 2X + 8 DX

解决方案：

Let us assume I =∫ cos2x/ √sin²2x+8 dx (i)

Put sin2x = t

2cos2x dx = dt

cos2x dx = dt/2

Put the above value in eq. (i)

=1/2 ∫ dt/ √t²+8

=1/2 ∫ dt/ √t²+(2√2)²

Integrate the above eq. then, we get

= 1/2 log|t +√t²+(2√2)²| + c [since ∫ 1/√x²+a² dx =log|x +√x²+a²| + c]

= 1/2 log|sin2x +√sin²2x+(2√2)²| + c

Hence, I = 1/2 log|sin2x +√sin²2x+8| + c

问题10.评估∫sin2x ^/√sin4×+ 4sin ² X-2 DX

解决方案：

Let us assume I =∫ sin2x/ √sin⁴x+4sin²x-2 dx (i)

Put sin²x = t

2sinxcosx dx = dt

sin2x dx = dt

Put the above value in eq. (i)

= ∫ dt/ √t²+4t-2

= ∫ dt/ √t²+2t(2)+(2)²-(2)²-2

= ∫ dt/ √(t+2)²-6 (ii)

Put t+2 =u

dt = du

Put the above value in eq. (ii)

= ∫ du/ √u²-6

= ∫ du/ √u²-(√6)²

Integrate the above eq. then, we get

= log|u +√u²-(√6)²| + c [since ∫ 1/√x²-a² dx =log|x +√x²-a²| + c]

= log|t+2 +√(t+2)²-6| + c

= log|sin²x+2 +√(sin²x+2)²-6| + c

= log|sin²x+2 +√sin⁴x+4sin²x+4-6| + c

Hence, I = log|sin²x+2 +√sin⁴x+4sin²x-2| + c