Question

Evaluate the given expression:

Answer 1

Answer:-

$\mathbf{1\ -\ \frac{\pi}{4}}$

Step-by-step explanation:

In this question,

We have to calculate

$\int\limits^\frac{\pi}{4}_0 {tan^{2}x} \, dx$

Therefore we know that,

$\mathbf{1\ +\ tan^{2}x\ =\ Sec^{2}x}$

$tan^{2}x\ =\ Sec^{2}x\ -\ 1$

Putting the value in integral we get,

$\int\limits^\frac{\pi}{4}_0 ({Sec^{2}x\ -\ 1}) \, dx$

Solving the integral we get,

$\int\limits^\frac{\pi}{4}_0 {Sec^{2}x}\ dx -\ \int\limits^\frac{\pi}{4}_0 {1}\ dx$

$\int\limits^\frac{\pi}{4}_0 {Sec^{2}x}\ dx\ =\ tanx$

Again Solving for upper and lower limits

$_{x=0}^{x=\frac{\pi}{4}}|tanx|\ -\ _{x=0}^{x=\frac{\pi}{4}}|x|$

On solving we get,

$\mathbf{1\ -\ \frac{\pi}{4}}$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

$\displaystyle \sf{\int\limits_{0}^{ \frac{\pi}{4} } { \tan}^{2}x \, dx }$

CALCULATION

$\displaystyle \sf{\int\limits_{0}^{ \frac{\pi}{4} } { \tan}^{2}x \, dx }$

$= \displaystyle \sf{\int\limits_{0}^{ \frac{\pi}{4} } ( { \sec}^{2}x - 1) \, dx }$

$= \displaystyle \sf{\int\limits_{0}^{ \frac{\pi}{4} } { \sec}^{2}x \, dx - \:\int\limits_{0}^{ \frac{\pi}{4} } \, dx \: }$

$= \displaystyle \sf{ \bigg [\: \tan x \: \bigg ]_0^{ \frac{\pi}{4} } \: - { \bigg [\: x \: \bigg ]_0^{ \frac{\pi}{4} }} \: }$

$= \displaystyle \sf{ ( \: \tan \frac{\pi}{4} - \tan \: 0) - ( \: \frac{\pi}{4} - 0 \: )\: }$

$= \displaystyle \sf{ 1 - 0 - \frac{\pi}{4} + 0\:}$

$= \displaystyle \sf{ 1 - \frac{\pi}{4} \:}$

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