Question

Find the integration of the following function

$\frac{sinx + cosx}{25 - 16sin2x}$
​

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{sinx + cosx}{25 - 16sin2x} \: dx$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{sinx + cosx}{9 + 16 - 16sin2x} \: dx$

$\rm \:  =  \: \displaystyle\int\rm \frac{sinx + cosx}{9 + 16(1 - sin2x)} \: dx$

$\rm \:  =  \: \displaystyle\int\rm \frac{sinx + cosx}{9 + 16( {sin}^{2}x + {cos}^{2}x- 2sinxcos)} \: dx$

$\rm \:  =  \: \displaystyle\int\rm \frac{sinx + cosx}{9 + 16( {sinx - cosx)}^{2}x } \: dx$

Now, to evaluate this integral, we use method of Substitution.

So, Substitute

$\purple{\rm :\longmapsto\:sinx - cosx = y}$

$\purple{\rm :\longmapsto\:(cosx + sinx)dx = dy}$

So, on substituting these values, we get

$\rm \:  =  \: \displaystyle\int\rm \dfrac{dy}{9 + 16 {y}^{2} }$

$\rm \:  =  \: 16\displaystyle\int\rm \dfrac{dy}{{y}^{2} + \dfrac{9}{16} }$

$\rm \:  =  \: 16\displaystyle\int\rm \dfrac{dy}{{y}^{2} + \bigg(\dfrac{3}{4}\bigg)^{2} }$

We know,

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

So, using this identity, we get

$\rm \:  =  \: 16 \times \dfrac{4}{3} \: {tan}^{ - 1} \dfrac{y}{ \dfrac{3}{4} } + c$

$\rm \:  =  \:\dfrac{64}{3} \: {tan}^{ - 1} \dfrac{4y}{3} + c$

$\rm \:  =  \:\dfrac{64}{3} \: {tan}^{ - 1} \dfrac{4(sinx - cosx)}{3} + c$

Hence,

$\boxed{\tt{ \rm \:\displaystyle\int\rm \frac{sinx + cosx}{25 - 16sin2x} dx =  \: \dfrac{64}{3} \: {tan}^{ - 1} \dfrac{4(sinx - cosx)}{3} + c}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

FORMULA USED :-

$\boxed{\tt{ {sin}^{2}x + {cos}^{2}x = 1}}$

$\boxed{\tt{ {x}^{2} - 2xy + {y}^{2} = {(x - y)}^{2} \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

ADDITIONAL INFORMATION

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

Answer 2

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{sinx + cosx}{25 - 16sin2x} \: dx$

can be rewritten as

$\rm \: = \: \displaystyle\int\rm \frac{sinx + cosx}{9 + 16 - 16sin2x} \: dx \\ \\ \rm \: = \: \displaystyle\int\rm \frac{sinx + cosx}{9 + 16(1 - sin2x)} \: dx \\ \\ \rm \: = \: \displaystyle\int\rm \frac{sinx + cosx}{9 + 16( {sin}^{2}x + {cos}^{2}x- 2sinxcos)} \: dx \\ \\ \rm \: = \: \displaystyle\int\rm \frac{sinx + cosx}{9 + 16( {sinx - cosx)}^{2}x } \: dx$

Now, to evaluate this integral, we use method of Substitution.

So, Substitute

$\purple{\rm :\longmapsto\:sinx - cosx = y} \\ \\ \purple{\rm :\longmapsto\:(cosx + sinx)dx = dy} \\ \\ So, \: \: \: on \: \: \: substituting \: \: \: these \: \: \: values \: \: \: , we \: \: \: get \\ \\ \rm \: = \: \displaystyle\int\rm \dfrac{dy}{9 + 16 {y}^{2} } \\ \\ \rm \: = \: 16\displaystyle\int\rm \dfrac{dy}{{y}^{2} + \dfrac{9}{16} } \\ \\ \rm \: = \: 16\displaystyle\int\rm \dfrac{dy}{{y}^{2} + \bigg(\dfrac{3}{4}\bigg)^{2} }$

We know,

$\begin{gathered} \purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ {x}^{2} + {a}^{2} } = \frac{1}{a} {tan}^{ - 1} \frac{x}{a} + c}}} \\ \end{gathered}$

So, using this identity, we get

$\rm \: = \: 16 \times \dfrac{4}{3} \: {tan}^{ - 1} \dfrac{y}{ \dfrac{3}{4} } + c \\ \\ \rm \: = \:\dfrac{64}{3} \: {tan}^{ - 1} \dfrac{4y}{3} + c \\ \\ \rm \: = \:\dfrac{64}{3} \: {tan}^{ - 1} \dfrac{4(sinx - cosx)}{3} + c$

Hence,

$\begin{gathered}\boxed{\tt{ \rm \:\displaystyle\int\rm \frac{sinx + cosx}{25 - 16sin2x} dx = \: \dfrac{64}{3} \: {tan}^{ - 1} \dfrac{4(sinx - cosx)}{3} + c}} \\ \end{gathered}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

FORMULA USED :-

$\begin{gathered}\boxed{\tt{ {sin}^{2}x + {cos}^{2}x = 1}} \\ \end{gathered} \\ \\ \begin{gathered}\boxed{\tt{ {x}^{2} - 2xy + {y}^{2} = {(x - y)}^{2} \: }} \\ \end{gathered}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬