Question

# $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}$ is equal to:A. $\pi$B. $\dfrac{\pi }{2}$C. $\dfrac{\pi }{3}$D. $\dfrac{\pi }{4}$

Hint: For$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}$ , use $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, then multiply and divide by $2$and simplify. After that, split the term and use $\sin \theta +\cos \theta =u$and apply the limits. Simplify it, you will get the answer.

Complete step by step solution:
We have to integrate,
$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}$

We know the identity, $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$,
So substituting above we get,
$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\dfrac{\sin \theta }{\cos \theta }}}$
Also, multiplying and dividing by $2$ we get,
$\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos \theta d\theta }{\cos \theta +\sin \theta }}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta }}$
So we can write,
$\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta }}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{(\cos \theta +\sin \theta +\cos \theta -\sin \theta )d\theta }{\cos \theta +\sin \theta }}$
Now, splitting we get,
$\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta }}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{d\theta +\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{(\cos \theta -\sin \theta )d\theta }{\cos \theta +\sin \theta }}}$
So let, $\sin \theta +\cos \theta =u$
Now differentiating both sides we get,
$(\cos \theta -\sin \theta )d\theta =du$
For $\theta =\dfrac{\pi }{2}$, $u=1$ and $\theta =0$, $u=1$
$\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta }}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{d\theta +\dfrac{1}{2}\int\limits_{1}^{1}{\dfrac{du}{u}}}$
Now we know that, $\int{\dfrac{1}{u}=\log u+c}$

$\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta }}=\dfrac{1}{2}\left[ \theta \right]_{0}^{\dfrac{\pi }{2}}+\dfrac{1}{2}\left[ \log u \right]_{1}^{1}$
Now, applying the limit we get,

$\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta }}=\dfrac{1}{2}\left( \dfrac{\pi }{2}-0 \right)+\dfrac{1}{2}\left( \log 1-\log 1 \right)=\dfrac{\pi }{4}+0=\dfrac{\pi }{4}$
We get,
$\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\cos \theta d\theta }{\cos \theta +\sin \theta }}=\dfrac{\pi }{4}$
We get the answer as option(D).

Note: Read the question carefully. You must be familiar with the concept of integration. Also, donâ€™t make silly mistakes. While simplifying, take care that no term is missing. Also, take care of signs. Most of the mistakes occur while simplifying so avoid it.