# $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}$ is equal to:

A. $\pi $

B. $\dfrac{\pi }{2}$

C. $\dfrac{\pi }{3}$

D. $\dfrac{\pi }{4}$

Last updated date: 16th Mar 2023

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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, don’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.

=\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, don’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.

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