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

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

 

 

Posted by

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