Question

$\bf{\orange{Hola\ Guys\ !}}$
$\blue{\sf Integral\ Challenge\ !}$
$\displaystyle \dagger\ \; \rm \green{\int \dfrac{1}{sin^6x}\ dx}$
$\mathbf{ . \qquad \pink{Thank\ you\ !} \qquad . }$

Answer 1

Question :

   $\displaystyle \sf \quad \red{\int \dfrac{1}{sin^6x}\ dx}$

Solution :

    $\displaystyle \sf \int \dfrac{1}{sin^6x}\ dx$

$\longrightarrow \quad \displaystyle \sf \int \left(\dfrac{1}{sin^2x}\right)^2 \dfrac{1}{sin^2x}\ dx$

Let's use substitution method , where

$\rm \implies \blue{u = cot\ x}$

$\rm \implies u = \dfrac{cos\ x}{sin\ x}$

$\rm \implies du = \dfrac{sin\ x(-sin\ x)-cosx(cos\ x)}{sin^2x}\ dx$

$\rm \implies du = \dfrac{-sin^2x-cos^2x}{sin^2x}\ dx$

$\rm \implies \orange{-du = \dfrac{1}{sin^2x}\ dx}$

and also ,

$\rm \implies \blue{u = cot\ x}$

$\rm \implies u = \dfrac{cos\ x}{sin\ x}$

$\rm \implies u^2 = \dfrac{cos^2 x}{sin^2 x}$

$\rm \implies u^2 = \dfrac{1-sin^2x}{sin^2 x}$

$\rm \implies u^2 = \dfrac{1}{sin^2 x}-1$

$\rm \implies \orange{u^2 +1= \dfrac{1}{sin^2 x}}$

Sub. in our main integral , we will get ,

$\longrightarrow \quad \displaystyle \sf \int \left(u^2+1\right)^2 (-du)$

$\longrightarrow \quad \displaystyle \sf - \int u^4+2u^2+1\ du$

$\longrightarrow \quad \displaystyle \sf - \left[ \dfrac{u^5}{5} + \dfrac{2u^3}{3} + u \right]+ c$

$\bullet \sf \quad \purple{u=cot\ x}$

$\longrightarrow \quad \displaystyle \sf \pink{- \left[ \dfrac{cot^5x}{5} + \dfrac{2cot^3x}{3} + cot\ x \right]+ c}$

Answer 2

Step-by-step explanation:

solution :

• please check the attached file