IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

layman

$\int \frac {\cos ^{3}{x} + \cos ^{5}{x}}{\sin ^{2}{x} + \sin ^{4}{x}}\,dx$

= $\int \frac {\cos ^{3}{x}\cdot(1 + \cos ^{2}{x})}{\sin ^{2}{x} + \sin ^{4}{x}}\,dx$

Put $\sin{x} = t$ obviously, $dt = \cos{x}\cdot dx$

= $\int \frac {(1 - t^{2})\cdot (2 - t^{2})}{t^{2}\cdot (1 + t^{2})}\,dt$

= $\int \frac {t^{4} - 3\cdot t^{2} + 2}{t^{2}\cdot (1 + t^{2})}\,dt$

= $\int 1 + (\frac {2 - 4\cdot t^{2}}{t^{2}\cdot (1 + t^{2})})\,dt$

Doing partial fractions we get,

= $\int 1 + \frac {2}{t^{2}} - \frac {6}{1 + t^{2}}\,dt$

= $t - \frac {2}{t} - 6\cdot \tan^{ - 1} {t} + C$

= $\sin{x} - 2\cdot cosec x - 6\cdot \tan^{ - 1} {\sin{x}} + C$

Ask & Discuss Questions with Community & Experts