Mathematics

# $\displaystyle \int \dfrac {\cos^{2}x}{\sin^{2}x }dx=$

$\dfrac {1}{3}\tan^{3}x+c$

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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
If $\int { \cfrac { xdx }{ \sqrt { 1+{ x }^{ 2 }+\sqrt { { \left( 1+{ x }^{ 2 } \right) }^{ 3 } } } } } =k\sqrt { 1+\sqrt { 1+{ x }^{ 2 } } } +C$ then k equals to

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
$\displaystyle \int _0^1 3x^2+2x dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Evaluate the following integrals:$\displaystyle \int \sqrt{4-9x^{2}}dx$
• A. $\dfrac{x}{2}.\sqrt{4-9x^{2}}+\dfrac12 \sin^{-1}\left ( \dfrac{3x}{2} \right )+C$
• B. $\dfrac{x}{2}.\sqrt{4-9x^{2}}+ \sin^{-1}\left ( \dfrac{3x}{2} \right )+C$
• C. none of these
• D. $\dfrac{x}{2}.\sqrt{4-9x^{2}}+\dfrac23 \sin^{-1}\left ( \dfrac{3x}{2} \right )+C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
If $\int _0^1 xdx = \dfrac {\pi}{4} - \dfrac {1}{2} ln 2$ then the value of definite integral $\int _0^1 \tan^{-1} (1-x+x^2) dx$ equals :
• A. $\dfrac {\pi}{4} + ln 2$
• B. $\dfrac {\pi}{4} - ln2$
• C. $2 ln 2$
• D. $ln2$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
If $A=\displaystyle \int_{0}^{1}{\dfrac{{e}^{t}}{1+t}}dt$ then $\displaystyle \int_{0}^{1}{{e}^{t}ln(1+t)}dt=$
• A. $e\ ln{2}+A$
• B. $Ae\ ln{2}$
• C. $A\ ln{2}$
• D. $e\ ln{2}-A$