Mathematics

# $\displaystyle \int x\sec x^{2}dx$ is equal to

$\displaystyle \frac{1}{2}\log \left ( \sec x^{2}+\tan x^{2} \right )+k$

##### SOLUTION
$\displaystyle \int x\sec x^{2}dx$

Put $x^2=t$
$\Rightarrow 2xdx=dt$

So,
$\displaystyle I=\frac{1}{2}\int \sec tdt$

$\displaystyle I=\frac{1}{2} \log|\sec t+\tan t|+k$

$\displaystyle I=\frac { 1 }{ 2 } \log |\sec { x^{ 2 } } +\tan x^{ 2 }|+k$

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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
Evaluate : $\int {{2^{{2^{{2^x}}}}}{2^{{2^x}}}{2^x}dx.}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \int x^{3}e^{x^{2}}dx=$
• A. $e^{x^{2}}(x^{2}-1)+c$
• B. $\dfrac{1}{2}e^{x^{2}}(x^{2}+1)+c$
• C. $e^{x^{2}}(x^{2}+1)+c$
• D. $\displaystyle \frac{1}{2}e^{x^{2}}(x^{2}-1)+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $\int_{0}^{f(x)} t^{2} d t=x \cos \pi x,$ then $f^{\prime}({9})$ is
• A. $-\dfrac{1}{3}$
• B. $\dfrac{1}{3}$
• C. non-existent
• D. $-\dfrac{1}{9}$

1 Verified Answer | Published on 17th 09, 2020

Q4 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. $\displaystyle \frac{1}{36} \tan^{-1} \left (\displaystyle \frac{4x^{3}}{3} \right) + c$

Let $\displaystyle f\left ( x \right )=\frac{\sin 2x \cdot \sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }$