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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 107 Maths Textbook Solution.

Provide Solution For R.D. Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 107  Maths Textbook Solution.

Answers (1)

Answer:\frac{1}{\sqrt{2}}\left[\frac{1}{2 \sqrt{\sqrt{2}+1}} \log \left|\frac{\sqrt{\sqrt{2}+1}+t}{\sqrt{\sqrt{2}+1}-t}\right|+\frac{1}{\sqrt{\sqrt{2}+1}} \tan ^{-1} \frac{t}{\sqrt{\sqrt{2}+1}}\right]+C

                                                                                                          \text { wheret }=\sin x-\cos x

Hint: to solve this we have to put \sin x+\cos u in team of t

Given:   \int \frac{\sin x+\cos x}{\sin ^{4} x+\cos ^{4} x} d x

Solution:

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

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

=\int \frac{\sin x+\cos x}{1-2 \sin ^{2} \cos ^{2} x} d x

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

=\int \frac{\sin x+ \cos x}{1-\frac{1}{2} \sin ^{2} 2 x} d x

\text { putting } \sin x-\cos x=t(i)

\Rightarrow(\sin x-\cos x)^{2}=t^{2}

\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x=t^{2}

1-2 \sin x \cos x=t^{2}

\sin 2 x=1-t^{2}

\text { Differentiating }(i), \text { weget }

(\cos x+\sin x) d x=d t

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

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

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

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

=\frac{2}{2 \sqrt{2}} \int\left[\frac{1}{\sqrt{2}+1-t^{2}}+\frac{1}{\sqrt{2}-1+t^{2}}\right] d t

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

=\frac{1}{\sqrt{2}} \int \frac{1}{(\sqrt{\sqrt{2}+1})^{2}-t^{2}} d t+\frac{1}{\sqrt{2}} \int \frac{1}{(\sqrt{\sqrt{2}-1})^{2}+t^{2}} d t

=\frac{1}{\sqrt{2}} \times \frac{1}{2 \sqrt{\sqrt{2}+1}} \log \left|\frac{\sqrt{\sqrt{2}+1}+t}{\sqrt{\sqrt{2}+1}-t}\right|+\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{\sqrt{2}+1}} \tan ^{-1} \frac{t}{\sqrt{\sqrt{2}+1}}+C

=\frac{1}{\sqrt{2}}\left[\frac{1}{2 \sqrt{\sqrt{2}+1}} \log \left|\frac{\sqrt{\sqrt{2}+1}+t}{\sqrt{\sqrt{2}+1}-t}\right|+\frac{1}{\sqrt{\sqrt{2}+1}} \tan ^{-1} \frac{t}{\sqrt{\sqrt{2}+1}}\right]+C

\text { wheret }=\sin x-\cos x

Posted by

infoexpert21

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