# Evaluate : $$\int$$ $$cos^4xdx\over {sin^3x{(sin^5x + cos^5x)^{3\over 5}}}$$

## Solution :

I = $$\int$$ $$cos^4xdx\over {sin^3x{(sin^5x + cos^5x)^{3\over 5}}}$$

= $$\int$$ $$cos^4xdx\over {sin^6x{(1 + cot^5x)^{3\over 5}}}$$ = $$\int$$ $$cot^4xcosec^2xdx\over {{(1 + cot^5x)^{3\over 5}}}$$

Put $$1+cot^5x$$ = t

$$5cot^4xcosec^2x$$dx = -dt

= -$$1\over 5$$ $$\int$$ $$dt\over {t^{3/5}}$$ = -$$1\over 2$$ $$t^{2/5}$$ + C

= -$$1\over 2$$ $${(1+cot^5x)}^{2/5}$$ + C

### Similar Questions

What is the integration of x tan inverse x dx ?

Prove that $$\int_{0}^{\pi/2}$$ log(sinx)dx = $$\int_{0}^{\pi/2}$$ log(cosx)dx = -$$\pi\over 2$$log 2.

What is the integration of tan inverse root x ?

Evaluate : $$\int$$ $$dx\over {3sinx + 4cosx}$$