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Question: \(\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}\) is equal to: A. \(\pi \) ...

∫0π2dθ1+tan⁡θ\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }} is equal to:
A. π\pi
B. π2\dfrac{\pi }{2}
C. π3\dfrac{\pi }{3}
D. π4\dfrac{\pi }{4}

Explanation

Solution

Hint: For∫0π2dθ1+tan⁡θ\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }} , use tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta }, then multiply and divide by 22and simplify. After that, split the term and use sin⁡θ+cos⁡θ=u\sin \theta +\cos \theta =uand apply the limits. Simplify it, you will get the answer.

Complete step by step solution:
We have to integrate,
∫0π2dθ1+tan⁡θ\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}

We know the identity, tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta },
So substituting above we get,
∫0π2dθ1+tan⁡θ=∫0π2dθ1+sin⁡θcos⁡θ\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 22 we get,
∫0π2dθ1+tan⁡θ=∫0π2cos⁡θdθcos⁡θ+sin⁡θ=12∫0π22cos⁡θdθcos⁡θ+sin⁡θ\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{(\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.