Mathematics

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


ANSWER

$$\pi/2$$


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
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