Mathematics

# Evaluate $\displaystyle \int \frac{\sec^{2}\:x}{\sqrt{\tan^{2}\:x+4}}dx.$

$\displaystyle=\log\left | \tan\:x+\sqrt{\tan^{2}\:x+4} \right |+C$

##### SOLUTION
Since derivative of $\displaystyle \tan \:x$ is $\displaystyle \sec^{2}\:x.$

Let $\displaystyle \tan\:x=t$ or $\displaystyle \sec^{2}x\:dx=dt$

$\displaystyle \therefore \int \frac{\sec^{2}\:x}{\sqrt{\tan^{2}\:x+4}}dx=\int \frac{dt}{\sqrt{t^{2}+2^{2}}}$

$\displaystyle=\log\left | t+\sqrt{t^{2}+4} \right |+C$

$\displaystyle=\log\left | \tan\:x+\sqrt{\tan^{2}\:x+4} \right |+C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
Evaluate $\displaystyle\int \frac{x^{2}}{9 + 16 x^{6}} dx$
• A. $\displaystyle \frac{1}{16} \tan^{-1} \left (\displaystyle \frac{4x^{3}}{3} \right) + c$
• B. $\displaystyle \frac{1}{36} \tan^{-1} \left (\displaystyle \frac{3x^{3}}{4} \right) + c$
• C. $\displaystyle \frac{1}{16} \tan^{-1} \left (\displaystyle \frac{3x^{3}}{4} \right) + c$
• D. $\displaystyle \frac{1}{36} \tan^{-1} \left (\displaystyle \frac{4x^{3}}{3} \right) + c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
If $\displaystyle I_{1}= \int_{-4}^{-5}e^{\left ( x +5 \right )^{2}}dx$ and $\displaystyle I_{2}= 3\int_{{1}/{3}}^{{2}/{3}}e^{\left ( 3x -2 \right )^{2}}dx$

then $I_{1}+I_{2}$ equals?
• A. $\displaystyle \frac{1}{3}$
• B. $\displaystyle -\frac{1}{3}$
• C. None of these
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Prove that $\displaystyle\int^{\pi}_0log(1+\cos x)dx=-\pi(log 2)$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate $\displaystyle\int^{1/2}_{1/4}\dfrac{dx}{\sqrt{x-x^2}}$.

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