Mathematics

# $\int_0^\pi {{x^2}\,g\left( x \right)\,dx\, = }$

$\frac{\pi }{8}$

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

#### Realted Questions

Q1 Single Correct Hard
$\displaystyle\int{\frac{(1+x^2)dx}{(1-x^2)\sqrt{1+x^2+x^4}}} =$
• A. $\displaystyle I=-\frac{1}{2\sqrt{3}}\log{\left|\frac{\sqrt{x^2+\displaystyle\frac{1}{x^2}+1}-\sqrt{5}}{\sqrt{x^2+\displaystyle\frac{1}{x^2}+1}+\sqrt{5}}\right|}+C$
• B. $\displaystyle I=-\frac{1}{4\sqrt{3}}\log{\left|\frac{\sqrt{x^2+\displaystyle\frac{1}{x^2}+1}-\sqrt{3}}{\sqrt{2x^2+\displaystyle\frac{1}{x^2}+1}+\sqrt{3}}\right|}+C$
• C. $\displaystyle I=-\frac{1}{2\sqrt{3}}\log{\left|\frac{\sqrt{2x^2+\displaystyle\frac{1}{2x^2}+1}-\sqrt{3}}{\sqrt{2x^2+\displaystyle\frac{1}{2x^2}+1}+\sqrt{3}}\right|}+C$
• D. $\displaystyle I=-\frac{1}{2\sqrt{3}}\log{\left|\frac{\sqrt{x^2+\displaystyle\frac{1}{x^2}+1}-\sqrt{3}}{\sqrt{x^2+\displaystyle\frac{1}{x^2}+1}+\sqrt{3}}\right|}+C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
$\int {cosec\;x dx}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate $\displaystyle \int_0^\pi {\frac{{{x^2}\sin 2x.\sin \left( {\frac{\pi }{2}\cos x} \right){\rm{ dx}}}}{{\left( {2x - \pi } \right)}}}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate $\int e^{x} \sec x (1+\tan x)dx$.

$\displaystyle \int_{0}^{\pi. }\frac{dx}{1+2\sin ^{2}x}= \frac{\pi }{\sqrt{\left ( k \right )}}$