Question

integrate (sin^4x) / (cos^6x) dx xnot equal to (2n+1)π/2,nbelongs toz​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{ {sin}^{4} x}{ {cos}^{6}x} \: dx$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{ {sin}^{4} x}{ {cos}^{4}x \times {cos}^{2} x} \: dx$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{ {sin}^{4} x}{ {cos}^{4}x} \times \frac{1}{ {cos}^{2}x} \: dx$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \frac{sinx}{cosx} = tanx}}} \: \\ \\ \purple{\rm :\longmapsto\:\boxed{\tt{ \frac{1}{cosx} = secx}}}$

So, on substituting these Identities, we get

$\rm \:  =  \: \displaystyle\int\rm {tan}^{4}x \: {sec}^{2}x \: dx$

Now,

To evaluate this integral, we use Method of Substitution

So, Substitute

$\purple{\rm :\longmapsto\:tanx = y} \: \: \: \: \: \: \: \: \: \: \: \\ \\ \purple{\rm :\longmapsto\: {sec}^{2} x \: dx \: = \: dy}$

So, on substituting these values, we get

$\rm \:  =  \: \displaystyle\int\rm {y}^{4} \: dy$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int\rm {x}^{n} \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: }}}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{ {y}^{4 + 1} }{4 + 1} + c$

$\rm \:  =  \: \dfrac{ {y}^{5} }{5} + c$

$\rm \:  =  \: \dfrac{ {tan}^{5} x}{5} + c$

Hence,

$\purple{\rm\implies \:\:\boxed{\tt{ \displaystyle\int\bf \frac{ {sin}^{4} x}{ {cos}^{6}x} \: dx \: = \: \frac{ {tan}^{5} x}{5} + c \: }}}$

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MORE TO KNOW

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Answer:

$\large\underline{\bf\red{Question-}}$

integrate (sin^4x) / (cos^6x) dx xnot equal to (2n+1)π/2,nbelongs toz

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}$

$\large\underline{\bf\red{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{ {sin}^{4} x}{ {cos}^{6}x} \: dx$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{ {sin}^{4} x}{ {cos}^{4}x \times {cos}^{2} x} \: dx$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{ {sin}^{4} x}{ {cos}^{4}x} \times \frac{1}{ {cos}^{2}x} \: dx$

We know,

$\green{\rm :\longmapsto\:\boxed{\tt{ \frac{sinx}{cosx} = tanx}}} \: \\ \\ \purple{\rm :\longmapsto\:\boxed{\tt{ \frac{1}{cosx} = secx}}}$

So, on substituting these Identities, we get

$\rm \:  =  \: \displaystyle\int\rm {tan}^{4}x \: {sec}^{2}x \: dx$

Now,

To evaluate this integral, we use Method of Substitution

So, Substitute

$\green{\rm :\longmapsto\:tanx = y} \: \: \: \: \: \: \: \: \: \: \: \\ \\ \purple{\rm :\longmapsto\: {sec}^{2} x \: dx \: = \: dy}$

So, on substituting these values, we get

$\rm \:  =  \: \displaystyle\int\rm {y}^{4} \: dy$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int\rm {x}^{n} \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: }}}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{ {y}^{4 + 1} }{4 + 1} + c$

$\rm \:  =  \: \dfrac{ {y}^{5} }{5} + c$

$\rm \:  =  \: \dfrac{ {tan}^{5} x}{5} + c$

Hence,

$\green{\rm\implies \:\:\boxed{\tt{ \displaystyle\int\bf \frac{ {sin}^{4} x}{ {cos}^{6}x} \: dx \: = \: \frac{ {tan}^{5} x}{5} + c \: }}}$

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