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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 76 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 76 Maths Textbook Solution.

Answers (1)

Answer:

-\frac{1}{3 \cos ^{8} x}-\cos x-\frac{2}{\cos x}+c

Hint:

To solve the given statement we will split sin 5x into sin? x sin x.

Given:

\int \frac{\sin ^{5} x}{\cos ^{4} x} d x

Solution:

=\int \frac{\sin ^{4} x \sin x}{\cos ^{4} x} d x

   =\int \frac{\left(1-\cos ^{2} x\right)^{2} \sin x}{\cos ^{4} x} d x \quad[\text { put } \cos x=t,-\sin x d x=d t]

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

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

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

=-\int \frac{1}{t^{4}} d t-\int d t+2 \int \frac{1}{t^{2}} d t

=-\frac{t^{-8}}{-3}-t+2\left(-\frac{t}{t}\right)+c

=-\frac{1}{3 \cos ^{8} x}-\cos x-\frac{2}{\cos x}+c

 

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