Question

$\displaystyle \huge \tt \color {green}\int_{0}^{\pi} \tan ^{ - 1} \bigg(20 {21}^{cosx} \bigg ) \: dx$​

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \tt{I = \int_{0}^{\pi} \tan ^{ - 1} \left( \left(2021 \right)^{cos(x)} \right) \: dx \: \: \: \: \: \: \: \: \: ...(1)}$

$\displaystyle \tt{ \implies \: I = \int_{0}^{\pi} \tan ^{ - 1} \left( \left(2021 \right)^{cos( \pi - x)} \right) \: dx}$

$\displaystyle \tt{ \implies \: I = \int_{0}^{\pi} \tan ^{ - 1} \left( \left(2021 \right)^{ - cos( x)} \right) \: dx}$

$\displaystyle \tt{ \implies \: I = \int_{0}^{\pi} \tan ^{ - 1} \left( \dfrac{1}{ \left(2021 \right)^{ - cos( x)}} \right) \: dx}$

$\displaystyle \tt{ \implies \: I = \int_{0}^{\pi} \cot ^{ - 1} \left( \left(2021 \right)^{ - cos( x)} \right) \: dx \: \: \: \: \: \: \: \: \: ...(2)}$

Adding (1) and (2),

$\displaystyle \tt{ \implies \:2 I =\int_{0}^{\pi} \tan^{ - 1} \left( \left(2021 \right)^{ - cos( x)} \right) \: dx + \int_{0}^{\pi} \cot ^{ - 1} \left( \left(2021 \right)^{ - cos( x)} \right) \: dx }$

$\displaystyle \tt{ \implies \:2 I =\int_{0}^{\pi} \left \{\tan^{ - 1} \left( \left(2021 \right)^{ - cos( x)} \right) + \cot ^{ - 1} \left( \left(2021 \right)^{ - cos( x)} \right) \right \}\: dx }$

$\displaystyle \tt{ \implies \:2 I =\int_{0}^{\pi} \dfrac{ \pi}{2}\: dx }$

$\displaystyle \tt{ \implies \:2 I = \dfrac{ \pi}{2}\:\int_{0}^{\pi} \: dx }$

$\displaystyle \tt{ \implies \:2 I = \dfrac{ \pi}{2} \cdot[x] _{0}^{\pi} }$

$\displaystyle \tt{ \implies \:2 I = \dfrac{ \pi}{2} (\pi - 0) }$

$\displaystyle \tt{ \implies \:2 I = \dfrac{ \pi^{2} }{2} }$

$\displaystyle \tt{ \implies \: I = \dfrac{ \pi^{2} }{4} }$