IIT-JEE 1995 ScreeningQuestion 1
The value of the integral $\displaystyle\int\frac{\cos^{3}x+\cos^{5}x}{\sin^{2}x+\sin^{4}x}\,dx$ is
Indefinite Integration
Solution
Concept. Take one $\cos x$ for $dt$ with $t=\sin x$; the rest is a rational function of $t$. Formulas used. (1) $t=\sin x$: $dt=\cos x\,dx$, $\cos^2x=1-t^2$(2) $\dfrac{-4t^2+2}{t^2(1+t^2)}=\dfrac{2}{t^2}-\dfrac{6}{1+t^2}$Step 1 - Substitute. $\cos^3x+\cos^5x=\cos x\big[(1-t^2)+(1-t^2)^2\big]$, so by (1) $$\int\frac{(1-t^2)+(1-t^2)^2}{t^2(1+t^2)}\,dt=\int\frac{t^4-3t^2+2}{t^4+t^2}\,dt.$$ Step 2 - Divide. $\frac{t^4-3t^2+2}{t^4+t^2}=1+\frac{-4t^2+2}{t^2(1+t^2)}$. By (2) (check: $2(1+t^2)-6t^2=2-4t^2$), the integrand is $1+\frac2{t^2}-\frac6{1+t^2}$. Step 3 - Integrate. $$t-\frac2t-6\tan^{-1}t+c=\sin x-2(\sin x)^{-1}-6\tan^{-1}(\sin x)+c.$$ Answer: (C)