Mathematics

# $\int (1- \cos x)\ \text{cosec}^2x \;dx = f(x) +c\; implies\; f(x) =$

$\tan \dfrac{x}{2}$+C

##### SOLUTION
$\int { \left( 1-\cos x \right) { \cos ec }^{ 2 }xdx }$
$\Rightarrow \int { { \cos ec }^{ 2 }xdx-\int { \dfrac { \cos x }{ { \sin }^{ 2 }x } dx } }$
$\sin x=m$
$\cos xdx=dm$
$\Rightarrow -cotx-\int { \dfrac { dm }{ { m }^{ 2 } } }$
$\Rightarrow -cotx+\dfrac { 1 }{ m }$
$\Rightarrow -cotx+\dfrac { 1 }{ \sin x }$
$\Rightarrow \dfrac { 1-\cos x }{ \sin x }$
$\Rightarrow \dfrac { 2{ \sin }^{ 2 }\left( x/2 \right) }{ 2\sin \left( \dfrac { x }{ 2 } \right) \cos \left( \dfrac { x }{ 2 } \right) } =\tan\left( x/2 \right) +c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

#### Realted Questions

Q1 Single Correct Medium
${\int}_{0}^{a}\left[f\left(a+x\right)+f\left(a-x\right)\right]dx=$
• A. ${2\int}_{0}^{a}f(x)dx$
• B. ${\int}_{0}^{2a}f(x)dx$
• C. ${\int}_{a}^{2a}f(x)dx$
• D. ${\int}_{-a}^{a}f(x)dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int\{\frac{(\log \mathrm{x}-1)}{(1+(\log \mathrm{x})^{2}}\}^{2}$ dx is equal to

• A. $\dfrac{logx}{(logx)^{2}+1}+c$
• B. $\dfrac{x}{x^{2}+1}+c$
• C. $\dfrac{xe^{x}}{1+x^{2}}+c$
• D. $\dfrac{x}{(logx)^{2}+1}+c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
$\text { Evaluate } \int_{0}^{\pi} e^{2 x} \cdot \sin \left(\dfrac{\pi}{4}+x\right) \mathrm{d} \mathrm{x}$

1 Verified Answer | Published on 17th 09, 2020

Q4 One Word Medium
Evaluate:$\displaystyle \int\frac{dx}{(4x^{2}-4x+3)}$

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