Mathematics

# The integral ${\int}_{\pi/12}^{\pi/4}\dfrac{8\cos 2x}{\left(\tan x+\cot x\right)^{3}}dx$ equals:

$\dfrac{15}{128}$

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

#### Realted Questions

Q1 Subjective Medium
Solve:
$\displaystyle \int \cfrac{x^{3}}{1+x^{4}} d x$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve $\displaystyle\int\dfrac {\sin x}{(1+\cos x)^{2}}dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int \dfrac {3^{x}}{\sqrt {1 - 9^{x}}} dx$ is equal to
• A. $(\log 3) \sin^{-1} (3^{x}) + C$
• B. $\dfrac {1}{3}\sin^{-1} (3^{x}) + C$
• C. $\dfrac {1}{9}\sin^{-1} (3^{x}) + C$
• D. $\sin^{-1} (3^{x}) + C$
• E. $\left (\dfrac {1}{\log 3}\right ) \sin^{-1}(3^{x}) + C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Find $\displaystyle \int \sqrt{10 - 4x + 4x^2} dx$

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 $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$