Question

Integration: 1/(3cosx-2sinx)^2

Answer 1

$\underline{\textsf{Question-}}$

$\textsf{Integrate:}\:\mathsf{\frac{1}{(3cosx-2sinx)^{2}}}\:\textsf{w. r. to x}$

$\underline{\textsf{Solution-}}$

$\textsf{Now,}\:\mathsf{\int \frac{dx}{(3cosx-2sinx)^{2}}}$

$\mathsf{=\int \frac{sec^{2}x\:dx}{(3-2tanx)^{2}}}$

$\textsf{Let, 3 - 2 tanx = z}$

$\mathsf{So,\: - 2 sec^{2}x\:dx = dz}$

$\mathsf{=-\frac{1}{2} \int \frac{dz}{z^{2}}}$

$\mathsf{=-\frac{1}{2}(-\frac{1}{z})+C}$

$\textsf{where C is integral constant}$

$\mathsf{=\frac{1}{2z}+C}$

$\mathsf{=\frac{1}{2(3-2tanx)}+C}$

$\mathsf{=\frac{1}{6-4tanx}+C}$

$\to \boxed{\tiny{\mathsf{\int \frac{dx}{(3cosx-2sinx)^{2}}=\frac{1}{6-4tanx}+C}}}$

$\underline{\textsf{Hence, solved.}}$

Answer 2

Answer:

Step-by-step explanation:

First,we divide numerator and denominator with cos^2(x),we convert integral into sec and tan form.

We substitute tanx=t,then apply another substitution 3-2t=u.

On solving,we get

1/{2(3-2tanx)}+c,where c is arbitrary constant