Mathematics

# $\int {\cos e{c^m}x\cot x\,dx = }$

$\frac{{ - 1}}{{m\,{{\sin }^m}x}} + C$

##### SOLUTION
$\begin{array}{l}I = \int {{\rm{cose}}{{\rm{c}}^m}x\cot xdx} \\I = \int {{\rm{cose}}{{\rm{c}}^{m - 2}}{\rm{cose}}{{\rm{c}}^2}x\cot xdx} \\{\rm{Let}},\\\cot x = t\\ - {{\mathop{\rm cosc}\nolimits} ^2}xdx = dt\\1 + {\cot ^2}x = {{\mathop{\rm cosc}\nolimits} ^2}x\\{\left( {1 + {t^2}} \right)^{\frac{1}{2}}} = {\mathop{\rm cosc}\nolimits} x\\I = \int { - {{\left( {1 + {t^2}} \right)}^{\frac{{m - 2}}{2}}}dt} \\ = \frac{{ - 1}}{{m{{\sin }^m}x}} + C\end{array}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

Q1 Single Correct Hard
lf $I_{m,n}=\displaystyle \int x^{m}(\log x)^{n}dx$, then$I_{m,n}-\displaystyle \frac{x^{m+1}}{(m+1)}(\log x)^{n}=$
• A. $\displaystyle \frac{n}{m+1},I_{mn-1}$
• B. $\displaystyle \frac{m}{n+1},I_{mn-1}$
• C. $\displaystyle \frac{n}{m+1}.I_{m-1}, n-1$
• D. $-\displaystyle \frac{n}{m+1},I_{mn-1}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve $\displaystyle\int {\dfrac{{\sin \theta \,d\theta }}{{\left( {4 + {{\cos }^2}\theta } \right)\left( {2 - {{\sin }^2}\theta } \right)}}}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Integrate the function    $\cfrac {1}{\sqrt {\sin^3x \sin (x+\alpha)}}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate:$\displaystyle \int_{1}^{2} \dfrac {x}{(x+1)(x+2)}dx$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
$\displaystyle \int \frac{cosec x}{\log \tan \left ( x/2 \right )}dx$
• A. $\displaystyle \log \left [ \log \tan \left ( x \right ) \right ].$
• B. $\displaystyle \log \left [ \log \cot \left ( x/2 \right ) \right ].$
• C. $\displaystyle \log \left [ \log \cot \left ( x \right ) \right ].$
• D. $\displaystyle \log \left [ \log \tan \left ( x/2 \right ) \right ].$