Mathematics

# If  $\displaystyle \int_{0}^{k}\frac{\cos x}{1+\sin^{2}x}dx=\frac{\pi}{4}$ then ${k}=?$

$\pi/2$

##### SOLUTION
Let $sin x = t$
Thus $cos x dx = dt$

Substituting the values back in the integral and performing indefinite integration,
$\displaystyle \int \dfrac{dt}{1 + {t}^{2}}$

$=$ ${tan}^{-1}t + c$

Substituting the value of t,
$=$ ${tan}^{-1}(sin x) + c$

Putting in the limits,
$=$ ${tan}^{-1}(sin k) - {tan}^{-1} 0 = \dfrac{\pi}{4}$

$=$ ${tan}^{-1}(sin k) - 0 = \dfrac{\pi}{4}$

$sin k = tan \dfrac{\pi}{4}$

$= sin k = 1$

or$k = \dfrac{\pi}{2}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

#### Realted Questions

Q1 Single Correct Medium
Evaluate the integral
$\displaystyle \int_{0}^{1}\tan^{-1}x \ dx$
• A. $\dfrac{\pi}{4}-\dfrac{1}{4} log2$
• B. $\displaystyle \frac{\pi}{4}+\frac{1}{2} log 2$
• C. $\displaystyle \frac{\pi}{4}+\frac{1}{4} log 2$
• D. $\displaystyle \frac{\pi}{4}-\frac{1}{2} log2$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate the following integrals
$\int { \cfrac { 1 }{ 3+4\cot { x } } } dx\quad$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate :
$\int\limits_0^{\pi /2} {\dfrac{{\cos x\,dx}}{{\left( {\,\cos \,x\, + \,\sin x} \right)}}}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
If $\alpha > 1$, then $\int \dfrac {dx}{x^{2} + 2\alpha x + 1} =$.
• A. $\dfrac {1}{\sqrt {1 - \alpha^{2}}}\cot^{-1} \left (\dfrac {x + \alpha}{\sqrt {1 - \alpha^{2}}}\right ) + c$
• B. $\dfrac {1}{2\sqrt {\alpha^{2} - 1}}\log \left (\dfrac {x + \alpha - \sqrt {\alpha^{2} - 1}}{x + \alpha + \sqrt {\alpha^{2} - 1}}\right ) + c$
• C. $\dfrac {1}{2\sqrt {\alpha^{2} - 1}}\log \left (\dfrac {x + \alpha + \sqrt {\alpha^{2} - 1}}{x + \alpha - \sqrt {\alpha^{2} - 1}}\right ) + c$
• D. $\dfrac {1}{\sqrt {1 - \alpha^{2}}}\tan^{-1} \left (\dfrac {x + \alpha}{\sqrt {1 - \alpha^{2}}}\right ) + c$

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