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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.

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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


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

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