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Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 77 Maths Textbook Solution.

Answers (1)

Answer:

I=\log (\sin x)-\sin ^{2} x+\frac{\sin ^{4} x}{4}+c

Hint:

to solve the given statement we will split cos? x into cos? x cos x then put cos? x = (1-sin²x)².

Given:

\int \frac{\cos ^{5} x}{\sin x} d x

Solution:

I=\frac{\cos ^{5} x}{\sin x} d x

    =\int \frac{\cos x\left(\cos ^{4} x\right)}{\sin x} d x

I=\int \frac{\cos x\left(1-\sin ^{2} x\right)^{2}}{\sin x} d x

\left[\because \sin x=t, \frac{d t}{d x}=\cos x, d x=\frac{d t}{\cos x}\right]

I=\int \frac{\left(1-t^{2}\right)^{2}}{t} d t

I=\int \frac{\left(1-2 t^{2}+t^{4}\right)}{t} d t

I=\int \frac{1}{t} d t-\int 2 t d t+\int t^{3} d t

I=\log t-t^{2}+\frac{t^{4}}{4}+c

I=\log (\sin x)-\sin ^{2} x+\frac{\sin ^{4} x}{4}+c

 

 

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