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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 72 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 72 Maths Textbook Solution.

Answers (1)

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Answer:

\frac{1}{2} \ln \left|\tan \frac{x}{2}\right|+\frac{1}{4} \tan ^{2} \frac{x}{2}+\tan \frac{x}{2}+C

Hint:

To solve the given statement multiply and divide the equation by sin x.

Given:

\int \frac{1+\sin x}{\sin x(1+\cos x)} d x

Solution:

I=\int \frac{(1+\sin x)}{\sin x(1+\cos x)} d x

\text { putting }

\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}

\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}

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

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

=\frac{1}{4} \int \frac{\left(1+\tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}\right) \sec ^{2} \frac{x}{2}}{\tan \frac{x}{2}} d x

\text { putting } \tan \frac{x}{2}=t

\Rightarrow \frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right) d x=d t

\Rightarrow \sec ^{2}\left(\frac{x}{2}\right) d x=2 d t

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

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

=\frac{1}{2}\left[\ln |t|+\frac{t^{2}}{2}+2 t\right]+C

=\frac{1}{2}\left[\ln \left|\tan \frac{x}{2}\right|+\frac{\tan ^{2}\left(\frac{x}{2}\right)}{2}+2 \tan \left(\frac{x}{2}\right)\right]+C\left[\because t=\tan \frac{x}{2}\right]

=\frac{1}{2} \ln \left|\tan \frac{x}{2}\right|+\frac{1}{4} \tan ^{2} \frac{x}{2}+\tan \frac{x}{2}+C

 

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