Question

Sina
dre
(05²2(4 1053 x=sinan)​

Answer 1

$\large\underline{\sf{Solution-}}$

The given integral

$\rm :\longmapsto\:\displaystyle\int\tt \sqrt{\dfrac{sinx}{ \: \: {cos}^{2} x(4 {cos}^{3}x - {sin}^{3}x) \: \: }} \: dx$

Identities Used:-

$\boxed{ \sf{ \: \frac{sinx}{cosx} = tanx}}$

$\boxed{ \sf{ \: \frac{1}{ {cos}x } = secx}}$

$\boxed{ \sf{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

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

$\boxed{ \sf{ \: \displaystyle\int\tt \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1} \frac{x}{a} + c}}$

Let's solve the problem now!!

$\rm :\longmapsto\:\displaystyle\int\tt \sqrt{\dfrac{sinx}{ \: \: {cos}^{2} x(4 {cos}^{3}x - {sin}^{3}x) \: \: }} \: dx$

$\rm \: = \:\displaystyle\int\tt \sqrt{\dfrac{sinx}{ \: \: {cos}^{2} x \: \times {cos}^{3}x(4 - {tan}^{3}x) \: \: }} \: dx$

can be rewritten as

$\rm \: = \:\displaystyle\int\tt \sqrt{\dfrac{sinx}{ \: \: {cos}x \: \times {cos}^{4}x(4 - {tan}^{3}x) \: \: }} \: dx$

$\rm \: = \:\displaystyle\int\tt \sqrt{\dfrac{tanx \times {sec}^{4} x}{ \: \: 4 - {tan}^{3}x \: \: }} \: dx$

$\rm \: = \:\displaystyle\int\tt \sqrt{\dfrac{tanx }{ \: \: 4 - {tan}^{3}x \: \: }} \: {sec}^{2} x \: dx$

$\rm \:  =  \:  \: \displaystyle\int\tt \dfrac{ \sqrt{tanx} \: \: {sec}^{2} x}{ \sqrt{4 - {tan}^{3} x} } \: dx$

Now, we use substitution method,

$\rm :\longmapsto\:Put \: {\bigg(tanx \bigg) }^{\dfrac{3}{2} }= y$

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

$\rm :\longmapsto\:\dfrac{3}{2} {(tanx)}^{ \dfrac{1}{2} }\dfrac{d}{dx}tanx = \dfrac{dy}{dx}$

$\rm :\longmapsto\:\dfrac{3}{2} \sqrt{tanx} {sec}^{2}x \: dx = dy$

$\rm :\longmapsto\: \sqrt{tanx} {sec}^{2}x \: dx = \dfrac{2}{3} dy$

On substituting all these values, we get

$\rm \:  =  \:  \: \dfrac{2}{3}\displaystyle\int\tt \dfrac{dy}{ \sqrt{4 - {y}^{2} } }$

$\rm \:  =  \:  \: \dfrac{2}{3}\displaystyle\int\tt \dfrac{dy}{ \sqrt{ {2}^{2} - {y}^{2} } }$

$\rm \:  =  \:  \: \dfrac{2}{3} {sin}^{ - 1}\dfrac{y}{2} + c$

$\rm \:  =  \:  \: \dfrac{2}{3} {sin}^{ - 1}\dfrac{ {( \sqrt{tanx}) }^{3} }{2} + c$

Hence,

$\rm :\longmapsto\:\displaystyle\int\tt \sqrt{\dfrac{sinx}{{cos}^{2} x(4 {cos}^{3}x - {sin}^{3}x)}} dx = \dfrac{2}{3} {sin}^{ - 1}\dfrac{ {( \sqrt{tanx}) }^{3} }{2} + c$

Additional Information :-

$\boxed{ \sf{ \: \displaystyle\int\tt \dfrac{dx}{ \sqrt{ {x}^{2} - {a}^{2} }} = log \: | \: x + \sqrt{ {x}^{2} - {a}^{2} } \: | + c}}$

$\boxed{ \sf{ \: \displaystyle\int\tt \dfrac{dx}{ \sqrt{ {x}^{2} + {a}^{2} }} = log \: | \: x + \sqrt{ {x}^{2} + {a}^{2} } \: | + c}}$

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