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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 78 Maths Textbook Solution.

Provide Solution For R.D. Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 78 Maths Textbook Solution.

Answers (1)

Answer:

\ln |\sec x+\tan x|-\sin x-\frac{\sin 3 x}{3}-\frac{\sin ^{5} x}{5}+c

Hint:

To solve the given equation we have to split the sin? x term in sin? x.sin² x.

Given:

\int \frac{\sin ^{6} x}{\cos x} d x

Solution:

I=\int \frac{\sin ^{6} x}{\cos x} d x

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

 =\int \frac{\sin ^{4} x}{\cos x} d x-\int \frac{\sin ^{4} x \cos ^{2} x}{\cos x} d x

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

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

 =\int \frac{\sin ^{2} x}{\cos x} d x-\int \sin ^{2} x \cos x d x-\frac{t^{5}}{5}

=\int \frac{\left(1-\cos ^{2} x\right)}{\cos x} d x-\int t^{2} d t-\frac{\sin ^{5} x}{5}

=\int \frac{1}{\cos x} d x-\int \cos x d x-\frac{t^{8}}{3}-\frac{\sin ^{5} x}{5}

=\int \sec x d x-\sin x-\frac{\sin 3 x}{3}-\frac{\sin ^{5} x}{5}

=\ln |\sec x+\tan x|+\left(-\sin x-\frac{\sin 3 x}{3}-\frac{\sin ^{5} x}{5}\right)+c

=\ln |\sec x+\tan x|-\sin x-\frac{\sin 3 x}{3}-\frac{\sin ^{5} x}{5}+c

 

Posted by

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