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

## Solution :

I = $$\int$$ $$dx\over {3sinx + 4cosx}$$ = $$\int$$ $$dx\over {3[{2tan{x\over 2}\over {1+tan^2{x\over 2}}}] + 4[{1-tan^2{x\over 2}\over {1+tan^2{x\over 2}}}]}$$ = $$\int$$ $$sec^2{x\over 2}dx\over {4+6tan{x\over 2}-4tan^2{x\over 2}}$$

let $$tan{x\over 2}$$ = t,

$$\therefore$$  $${1\over 2}sec^2{x\over 2}$$dx = dt

so I = $$\int$$ $$2dt\over {4+6t-4t^2}$$ = $$1\over 2$$ $$\int$$ $$dt\over {1-(t^2-{3\over 2}t})$$ = $$1\over 2$$ $$\int$$ $$dt\over {{25\over 16}-{(t-{3\over 4})}^2}$$

= $$1\over 2$$ $$1\over {2({5\over 4})}$$ $$ln|{{{5\over 4}+(t-{3\over 4})}\over {{5\over 4}-(t-{3\over 4})}}|$$ + C = $$1\over 5$$ $$ln|{1+2tan{x\over 2}\over {4-2tan{x\over 2}}}|$$ + 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.

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

What is the integration of tan inverse root x ?