Mathematics

# Solve:$\displaystyle\int_{0}^{\pi/4}\dfrac{\tan^{3}x}{1+\cos 2x}dx$

##### SOLUTION

Given $\displaystyle\int_{0}^{\pi/4}\dfrac{\tan^{3}x}{1+\cos 2x}dx$

$I=\displaystyle\int_{0}^{\pi/4}\dfrac{\tan^{3}x}{2\cos^{2}x}dx$    [$\because 2\cos^{2}x=1+\cos 2x$]

$=\dfrac{1}{2}\displaystyle\int_{0}^{\pi/4}\tan^{3}x\sec^{2}x\ dx$

$=\dfrac{1}{2}\displaystyle\int_{0}^{1}t^{3}dt$, where $t=\tan x$

$\Rightarrow I=\dfrac{1}{2}\left[\dfrac{t^{4}}{4}\right]_{0}^{1}$

$=\dfrac{1}{2}\left(\dfrac{1}{4}-0\right)$

$=\dfrac{1}{8}$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Subjective Medium
Evaluate the following as limit of sums:
$\displaystyle \int_{0}^{2} (x^2 + 3) dx$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate:
$\displaystyle\int{\dfrac{{x}^{2}}{1+{x}^{3}}}dx$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
The value of $\displaystyle \int ^{\tan x}_{1/e}\displaystyle \frac{t\, dt}{1+t^{2}}+\displaystyle \int ^{\cot x}_{1/e}\displaystyle \frac{dt}{t\left ( 1+t^{2} \right )}$ is
• A. $\displaystyle \frac{1}{2+tan^{2}x}$
• B. $\pi /4$
• C. $\displaystyle \frac{2}{\pi }\displaystyle \int ^{1}_{-1}\displaystyle \frac{dt}{1+t^{2}}$
• D. $1$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
The value of $\displaystyle\int\limits_{0}^{\frac{\pi}{4}} \tan^2 \theta\ d\theta=$
• A. $\dfrac{\pi}{4}-1$
• B. $\dfrac{\pi}{4}$
• C. none of these
• D. $1-\dfrac{\pi}{4}$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020