Question

Find the value of $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}$.
A. $\pi $
B. $\dfrac{\pi }{2}$
C. $\dfrac{\pi }{3}$
D. $\dfrac{\pi }{4}$

Answer 1

Hint:

• • In this question, we have $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}$. So, we will consider this as equation 1.

• Next, we will use the identity $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$ and then we will use the Cofunction identity of $\tan \left( \dfrac{\pi }{2}-\theta \right)$ is equal to $\cot \theta $.

• From this, we will obtain equation 2.

• Now we will add equations 1 and 2 and after that, we will split $\text{ }\cot \theta $ as $\dfrac{\cos \theta }{\sin \theta }$ and $\tan \theta $ as $\dfrac{\sin \theta }{\cos \theta }$.

• We will further take the LCM to get the simplified form of the given expression. At last, we will obtain the final result by substituting the values of the limits.

Formula Used: We will use the following formulas:
1) $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$
2) $\tan \left( \dfrac{\pi }{2}-\theta \right)=\cot \theta $
3) $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$

Complete step by step solution: In this question, we are given:
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}$ ….... (1)
First, we will use the identity $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \left( \dfrac{\pi }{2}-\theta \right)}}$
By using the Cofunction identity of $\tan \left( \dfrac{\pi }{2}-x \right)=\cot x$, we get
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\cot \theta }}$ ….... (2)
We will now add equations (1) and (2) to get.
$\begin{align}
  & I+I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\tan \theta }}+\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{d\theta }{1+\cot \theta }} \\
 & \Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{1}{1+\tan \theta }+\dfrac{1}{1+\cot \theta } \right)}d\theta \\
\end{align}$
Next, we will write $\cot \theta $ as $\dfrac{\cos \theta }{\sin \theta }$ and $\tan \theta $ as $\dfrac{\sin \theta }{\cos \theta }$.
$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{1}{1+\dfrac{\sin \theta }{\cos \theta }}+\dfrac{1}{1+\dfrac{\cos \theta }{\sin \theta }} \right)}d\theta $
Take LCM in the denominator to get
$\begin{align}
  & \Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{\cos \theta }{\cos \theta +\sin \theta }+\dfrac{\sin \theta }{\sin \theta +\cos \theta } \right)}d\theta \\
 & \Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{\cos \theta +\sin \theta }{\cos \theta +\sin \theta } \right)}d\theta \\
 & \Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{(1)d\theta } \\
\end{align}$
We will further integrate 1 with respect to $\theta $.
$\Rightarrow 2I=\left[ \theta \right]_{0}^{\dfrac{\pi }{2}}$
At last, substitute the values of the limit to get
$\begin{align}
  & \Rightarrow 2I=\left[ \dfrac{\pi }{2}-0 \right] \\
 & \Rightarrow I=\dfrac{\pi }{2\times 2} \\
 & \therefore I=\dfrac{\pi }{4} \\
\end{align}$

Option ‘D’ is correct

Note: As the functions change, it is not essential to integrate taking limits directly $0$ to $\dfrac{\pi }{2}$. We must avoid errors and confusion while dealing with sign changes that occurred during integration. We need to adopt the identities that make the problems easier.