Question

the question is from definite integrals​

Answer 1

$\rm \longrightarrow \displaystyle \int_{0}^{ \rm \frac{\pi}{2} } \rm \frac{1}{4 + 5 \: cos \: x} dx$

We know that :-

$\rm \: cos \: 2( \frac{x}{2} ) = cos \: x \\ \\ \boxed{ \rm cos \: 2( \frac{x}{2} ) = \dfrac{1 - {tan}^{2} \frac{x}{2} }{1 + {tan}^{2} \frac{x}{2} } }$

$\rm \longrightarrow \displaystyle \int_{0}^{ \rm \frac{\pi}{2} } \rm \frac{1}{4 + 5 \bigg(\dfrac{1 - {tan}^{2} \frac{x}{2} }{1 + {tan}^{2} \frac{x}{2} } \bigg)} dx$

$\rm \longrightarrow \displaystyle \int_{0}^{ \rm \frac{\pi}{2} } \rm \frac{1 + {tan}^{2} \frac{x}{2}}{4 (1 + {tan}^{2} \frac{x}{2})+ 5(1 - {tan}^{2} \frac{x}{2} )} dx$

$\rm \longrightarrow \displaystyle \int_{0}^{ \rm \frac{\pi}{2} } \rm \frac{1 + {tan}^{2} \frac{x}{2}}{4 (1 + {tan}^{2} \frac{x}{2})+ 5(1 - {tan}^{2} \frac{x}{2} )} dx$

$\rm \longrightarrow \displaystyle \int_{0}^{ \rm \frac{\pi}{4} } \rm \frac{1 + {tan}^{2} \frac{x}{2}}{4 + 4 \: {tan}^{2} \frac{x}{2}+ 5 - 5 \: {tan}^{2} \frac{x}{2} } dx$

$\rm \longrightarrow \displaystyle \int_{0}^{ \rm \frac{\pi}{2} } \rm \frac{1 + {tan}^{2} \frac{x}{2}}{9 - \: {tan}^{2} \frac{x}{2} } dx$

$\rm \longrightarrow \displaystyle \int_{0}^{ \rm \frac{\pi}{2} } \rm \dfrac{ {sec}^{2} \frac{x}{2} }{9 - \: {tan}^{2} \frac{x}{2} } dx$

$\rm let \: \: \: {tan} \: \frac{x}{2} = t \\ \\ \rm \frac{1}{2} \times {sec}^{2} \frac{x}{2} \: dx = dt\\ \\ \rm {sec}^{2} \frac{x}{2} \: dx =2 dt$

$\rm \: if \: x = \frac{\pi}{2} \: \: then : \\ \\ \: \rm tan \frac{\pi}{4} = 1 = t \\ \\ \rm \: if \: x = 0 \: \: then : \\ \\ \: \rm tan 0= 0= t$

$\rm \longrightarrow \displaystyle \int_{0}^{ \rm 1} \rm \dfrac{ 2 \: dt}{9 - \: {t}^{2} }$

$\rm \longrightarrow 2\displaystyle \int_{0}^{ \rm 1 } \rm \dfrac{ \: dt}{ {3}^{2} - \: {t}^{2} }$

$\rm \longrightarrow 2\displaystyle \bigg[ \rm\frac{1}{6} \: log \: \bigg | \dfrac{ 3 + t}{ {3} - \: {t} } \bigg |\bigg ]_{0}^{ \rm1}$

$\rm \longrightarrow \displaystyle \rm\frac{1}{3} \bigg( log \: \bigg | \dfrac{ 3 + 1}{ {3} - \: 1 } \bigg | - log \: \bigg | \dfrac{ 3 + 0}{ {3} - \: 0 } \bigg |\bigg)$

$\rm \longrightarrow \displaystyle \rm\frac{1}{3} \bigg( log \: \dfrac{ 4}{ 2} - log \: \dfrac{ 3}{ {3} } \bigg)$

$\rm \longrightarrow \displaystyle \rm\frac{1}{3} \bigg( log \: 2 - log \: 1 \bigg)$

$\rm \longrightarrow \displaystyle \rm\frac{1}{3} \bigg( log \: 2 - 0 \bigg)$

$\rm \longrightarrow \displaystyle \rm\frac{log \: 2}{3}$