Question

find the integral of : sin^6x/cos^8x ​

Answer 1

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{\red{\sf\:\displaystyle \int \sf \: {x}^{n} = \dfrac{ {x}^{n + 1} }{n + 1} + c}}$

$\boxed{\red{\sf\:\dfrac{d}{dx} tanx = {sec}^{2}x}}$

Let's solve the problem now!!

$\rm :\longmapsto\:Let \: A \: = \: \displaystyle \int \sf \: \dfrac{ {sin}^{6}x }{ {cos}^{8}x} dx$

$\: \sf \: \: = \: \: \displaystyle \int \sf \: \dfrac{ {sin}^{6}x}{ {cos}^{6}x \: \times \: {cos}^{2} x} \: dx$

$\: \sf \: \: = \: \: \displaystyle \int \sf \: {tan}^{6}x \: {sec}^{2}x \: dx$

Now, we use the method of Substitution, to solve this integral.

$\rm :\longmapsto\:\blue{ \sf \:Put \: tanx = y}$

On differentiating w. r. t. x, we get

$\rm :\longmapsto\:\blue{ \sf \: {sec}^{2} x \: dx \: = \: dy}$

On substituting all these values in integral, we get

$\rm :\longmapsto\:A \: = \: \displaystyle \int \sf \: {y}^{6} \: dy$

$\: \sf \: \: = \: \: \dfrac{ {y}^{6 + 1} }{6 + 1} + c$

$\: \sf \: \: = \: \: \dfrac{ {y}^{7} }{7} + c$

$\: \sf \: \: = \: \: \dfrac{ {tan}^{7}x}{7} + c$

Additional Information :-

$\boxed{\red{\sf\:\displaystyle \int \sf \: cosx \: dx = sinx \: + \: c}}$

$\boxed{\red{\sf\:\displaystyle \int \sf \: sinx \: dx = - \: cosx \: + \: c}}$

$\boxed{\red{\sf\:\displaystyle \int \sf \: tanx \: dx = - \:log(cosx) \: + \: c}}$

$\boxed{\red{\sf\:\displaystyle \int \sf \: cotx \: dx = \:log(sinx) \: + \: c}}$

$\boxed{\red{\sf\:\displaystyle \int \sf \: {sec}^{2}x = tanx + c}}$

$\boxed{\red{\sf\:\displaystyle \int \sf \: {cosec}^{2}x = - \: cotx + c}}$

$\boxed{\red{\sf\:\displaystyle \int \sf \: \dfrac{dx}{ {x}^{2} + 1} = {tan}^{ - 1}x + c}}$