Definite Integrals: Two Questions Using Properties

This lesson solves two important CBSE definite integral problems using properties of definite integrals. Both rely on the reflection property, which replaces $x$ by $a+b-x$ over the interval $[a,b]$, turning a hard integrand into something that combines neatly with the original.

Question 1: evaluate $\displaystyle\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\tan x}}$

Write $\tan x = \dfrac{\sin x}{\cos x}$ and simplify the denominator.

$I = \int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\tfrac{\sin x}{\cos x}}} = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx \quad (1)$

Apply the reflection property. Here $a+b = \tfrac{\pi}{6}+\tfrac{\pi}{3} = \tfrac{\pi}{2}$, so replace $x$ by $\tfrac{\pi}{2}-x$. Using $\cos\!\left(\tfrac{\pi}{2}-x\right)=\sin x$ and $\sin\!\left(\tfrac{\pi}{2}-x\right)=\cos x$:

$I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx \quad (2)$

Add (1) and (2). The numerators sum to the common denominator:

$2I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx = \int_{\pi/6}^{\pi/3} 1\,dx = \frac{\pi}{3}-\frac{\pi}{6} = \frac{\pi}{6}$

Therefore

$I = \frac{\pi}{12}.$

Question 2: prove $\displaystyle\int_{0}^{\pi/2} \log(\sin x)\,dx = -\frac{\pi}{2}\log 2$

This is a standard result worth memorising, as it reappears in other problems. Let

$I = \int_{0}^{\pi/2} \log(\sin x)\,dx \quad (1)$

Reflect. Replacing $x$ by $\tfrac{\pi}{2}-x$ and using $\sin\!\left(\tfrac{\pi}{2}-x\right)=\cos x$:

$I = \int_{0}^{\pi/2} \log(\cos x)\,dx \quad (2)$

Add (1) and (2).

$2I = \int_{0}^{\pi/2} \big(\log\sin x + \log\cos x\big)\,dx = \int_{0}^{\pi/2} \log(\sin x\cos x)\,dx$

Add and subtract $\log 2$ to introduce the double angle. Since $2\sin x\cos x = \sin 2x$:

$2I = \int_{0}^{\pi/2} \log\!\big(2\sin x\cos x\big)\,dx - \int_{0}^{\pi/2} \log 2\,dx = \int_{0}^{\pi/2} \log(\sin 2x)\,dx - \int_{0}^{\pi/2} \log 2\,dx$

Call these $I_1$ and $I_2$.

$2I = I_1 - I_2 \quad (3)$

Evaluate $I_1$

Put $2x = t$, so $dx = \tfrac{1}{2}\,dt$; the limits become $0$ to $\pi$.

$I_1 = \int_{0}^{\pi/2} \log(\sin 2x)\,dx = \frac{1}{2}\int_{0}^{\pi} \log(\sin t)\,dt$

Using $\displaystyle\int_{0}^{2a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx$ when $f(2a-x)=f(x)$, and $\sin(\pi-t)=\sin t$:

$I_1 = \frac{1}{2}\cdot 2\int_{0}^{\pi/2} \log(\sin t)\,dt = \int_{0}^{\pi/2} \log(\sin x)\,dx = I \quad (4)$

Evaluate $I_2$

Since $\log 2$ is constant,

$I_2 = \int_{0}^{\pi/2} \log 2\,dx = \log 2 \cdot \frac{\pi}{2} = \frac{\pi}{2}\log 2 \quad (5)$

Combine

Substituting (4) and (5) into (3):

$2I = I - \frac{\pi}{2}\log 2 \;\Rightarrow\; I = -\frac{\pi}{2}\log 2.$

Key takeaways

• The reflection property $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$ turns many symmetric integrals into a constant after adding the two forms.
• For $\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\tan x}}$, the two forms add to the integral of $1$, giving $\tfrac{\pi}{12}$.
• The result $\int_{0}^{\pi/2} \log(\sin x)\,dx = -\tfrac{\pi}{2}\log 2$ is a standard formula to memorise for later problems.