Question

solve the question step by step​

Answer 1

Let, I=∫(sinx−2cosx)(2sinx+cosx)1dx=∫2sin2x−3sinxcosx−2cos2x1dx=∫−2(cos2x−sin2x)−3sinxcosx1dx=−∫2cos2x+23sin2x1dx

Now, let tanx=t∴dx=1+t22dt,cos2x=1+t21−t2,sin2x=1+t22t

We can write, 

I=−∫2(1+t21−t2)+23(1+t2

Answer 2

Solution:

Given Integral:

$\displaystyle \rm \longrightarrow I = \int \dfrac{dx}{( \sin x - 2 \cos x)(2 \sin x + \cos x)}$

Can be written as:

$\displaystyle \rm \longrightarrow I = \int \dfrac{\sec^{2}x}{\sec^{2}x( \sin x - 2 \cos x)(2 \sin x + \cos x)} \: dx$

$\displaystyle \rm \longrightarrow I = \int \dfrac{\sec^{2} x}{\sec x( \sin x - 2 \cos x) \times \sec x(2 \sin x + \cos x)} \: dx$

$\displaystyle \rm \longrightarrow I = \int \dfrac{\sec^{2} x}{ \dfrac{1}{ \cos x} ( \sin x - 2 \cos x) \times \dfrac{1}{ \cos x} (2 \sin x + \cos x)} \: dx$

$\displaystyle \rm \longrightarrow I = \int \dfrac{\sec^{2} x}{( \tan x - 2)(2 \tan x + 1)} \: dx$

Now, let us assume that:

$\rm \longrightarrow u = \tan x$

$\rm \longrightarrow du= \sec^{2}x \: dx$

So, the integral changes to:

$\displaystyle \rm \longrightarrow I = \int \dfrac{du}{(u- 2)(2u + 1)}$

After converting to partial fraction, we get:

$\displaystyle \rm \longrightarrow I = \int \bigg[ \dfrac{1}{5(u - 2)} - \dfrac{2}{5(2u + 1)} \bigg] \: du$

$\displaystyle \rm \longrightarrow I = \dfrac{1}{5} \int \bigg[ \dfrac{1}{u - 2} - \dfrac{2}{2u + 1} \bigg] \: du$

$\displaystyle \rm \longrightarrow I = \dfrac{1}{5} \int \dfrac{du}{u - 2} - \dfrac{1}{5} \int \dfrac{2 \: du}{2u + 1}$

$\displaystyle \rm \longrightarrow I = \dfrac{1}{5} \int \dfrac{du}{u - 2} - \dfrac{2}{5} \int \dfrac{du}{2u + 1}$

Now, let us assume that:

$\rm \longrightarrow v=u - 1$

$\rm \longrightarrow dv=du$

$\rm \longrightarrow w=2u + 1$

$\rm \longrightarrow dw=2 \: du$

So, the integral changes to:

$\displaystyle \rm \longrightarrow I = \dfrac{1}{5} \int \dfrac{dv}{v} - \dfrac{2}{5} \int \dfrac{dw}{2w}$

$\displaystyle \rm \longrightarrow I = \dfrac{1}{5} \int \dfrac{dv}{v} - \dfrac{1}{5} \int \dfrac{dw}{w}$

$\displaystyle \rm \longrightarrow I = \dfrac{1}{5} \ln( |v| ) - \dfrac{1}{5} \ln( |w| ) + C$

$\displaystyle \rm \longrightarrow I = \dfrac{1}{5} \ln \bigg( \bigg| \dfrac{v}{w} \bigg | \bigg) + C$

Substituting back the values, we get:

$\displaystyle \rm \longrightarrow I = \dfrac{1}{5} \ln \bigg( \bigg| \dfrac{u - 2}{2u + 1} \bigg | \bigg) + C$

$\displaystyle \rm \longrightarrow I = \dfrac{1}{5} \ln \bigg( \bigg| \dfrac{ \tan x- 2}{2 \tan x+ 1} \bigg | \bigg) + C$

★ Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$