Question

$\bf{\orange{Hola\ Guys\ !}}$ $\blue{\sf Integral\ Challenge\ !}$ $\displaystyle \dagger\ \; \rm \green{\int \dfrac{1}{sin^6x}\ dx}$ ​

Answer 1

Answer:

$-\frac{1}{5} cot^{5}( x)-\frac{2}{3} cot^3x-cot(x)+C$

Explanation:

$\int\limits {\frac{1}{sin^{6}x } } \, dx$  $=-\int\limits ({\frac{1}{sin^{2}x } } )^2.\frac{-1}{sin^{2}x } \, dx$

Now,

$u=cot(x)$

$du=\frac{-1}{sin^{2}(x) }dx$

Also,

$u=cot(x)$

$u=\frac{cos(x)}{sin(x)}$

$u^{2} =\frac{cos^2x }{sin^{2}x}\\\\u^{2} =\frac{1-sin^{2}x}{sin^{2}x} \\\\u^{2} =\frac{1}{sin^{2}x}-1\\\\u^2+1=\frac{1}{sin^{2}x}$

Putting these values we get,

$=> -\int\limits( {u^2+1})^2 \, du$

$=> -\int\limits {(u^2+2u^2+1)} \, du$

$=> -[\frac{u^5}{5} +\frac{2u^3}{3} +u]$

$=> -\frac{1}{5} cot^{5}( x)-\frac{2}{3} cot^3x-cot(x)+C$

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Answer 2

Answer:

Answer:

-\frac{1}{5} cot^{5}( x)-\frac{2}{3} cot^3x-cot(x)+C−51cot5(x)−32cot3x−cot(x)+C

Explanation:

\int\limits {\frac{1}{sin^{6}x } } \, dx∫sin6x1dx  =-\int\limits ({\frac{1}{sin^{2}x } } )^2.\frac{-1}{sin^{2}x } \, dx=−∫(sin2x1)2.sin2x−1dx

Now,

u=cot(x)u=cot(x)

du=\frac{-1}{sin^{2}(x) }dxdu=sin2(x)−1dx

Also,

u=cot(x)u=cot(x)

u=\frac{cos(x)}{sin(x)}u=sin(x)cos(x)

\begin{gathered}u^{2} =\frac{cos^2x }{sin^{2}x}\\\\u^{2} =\frac{1-sin^{2}x}{sin^{2}x} \\\\u^{2} =\frac{1}{sin^{2}x}-1\\\\u^2+1=\frac{1}{sin^{2}x}\end{gathered}u2=sin2xcos2xu2=sin2x1−sin2xu2=sin2x1−1u2+1=sin2x1

Putting these values we get,

= > -\int\limits( {u^2+1})^2 \, du=>−∫(u2+1)2du

= > -\int\limits {(u^2+2u^2+1)} \, du=>−∫(u2+2u2+1)du

= > -[\frac{u^5}{5} +\frac{2u^3}{3} +u]=>−[5u5+32u3+u]

= > -\frac{1}{5} cot^{5}( x)-\frac{2}{3} cot^3x-cot(x)+C=>−51cot5(x)−32cot3x−cot(x)+C

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