Evaluate the definite integral: $\int_{0}^{\frac{\pi}{4}}\left(2 \sec ^{2} x+x^{3}+2\right) d x$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
Let $I = \int_{0}^{\frac{\pi}{4}} (2 \sec^2 x + x^3 + 2) dx$.
First,find the indefinite integral:
$\int (2 \sec^2 x + x^3 + 2) dx = 2 \tan x + \frac{x^4}{4} + 2x = F(x)$.
By the second fundamental theorem of calculus,$I = F\left(\frac{\pi}{4}\right) - F(0)$.
$F\left(\frac{\pi}{4}\right) = 2 \tan\left(\frac{\pi}{4}\right) + \frac{1}{4}\left(\frac{\pi}{4}\right)^4 + 2\left(\frac{\pi}{4}\right) = 2(1) + \frac{\pi^4}{4 \times 256} + \frac{\pi}{2} = 2 + \frac{\pi^4}{1024} + \frac{\pi}{2}$.
$F(0) = 2 \tan(0) + \frac{0^4}{4} + 2(0) = 0 + 0 + 0 = 0$.
Therefore,$I = 2 + \frac{\pi}{2} + \frac{\pi^4}{1024}$.

Explore More

Similar Questions

If $[\cdot]$ denotes the greatest integer function,then $\int_1^2 [x^2] dx =$

$\int_{e^{-1}}^{e^2} \left| \frac{\log x}{x} \right| dx =$

If $\frac{d}{dx}\{f(x)\} = g(x)$,then $\int_a^b f(x) g(x) dx$ is equal to

Suppose $g(x) = \int_0^x f(t) dt$,where $f$ is such that for $t \in [0, 1]$,$0 \le f(t) \le \frac{1}{2}$ and for $t \in [1, 2]$,$\frac{1}{2} \le f(t) \le 1$. Find the range of $g(2)$.

Difficult
View Solution

$\int_0^\pi \frac{1}{1+\sin x} dx$ is equal to

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo