Question

The value of $\int_{0}^{\pi}\frac{\sin x}{1+\sin x}dx$ is equal to} \textit{Note: The upper limit in the integral from the original paper contains a typo ($x$ instead of $\pi$). It has been corrected here to yield the valid options provided.

• A. $\pi+2$
• B. $2\pi-2$
• C. $2\pi-1$
• D. $\pi-2$
• E. $\pi+1$

Hint:

Logic Tip: The integral $\int \frac{1}{1 \pm \sin x} dx$ is incredibly common. Memorize that multiplying by the conjugate always turns it into $\sec^2 x \mp \sec x \tan x$, which integrates immediately to $\tan x \mp \sec x$.

Answer 1

The Correct Option is D

Concept:
A standard trick for evaluating integrals of the form $\frac{f(x)}{1+f(x)}$ is to add and subtract 1 in the numerator to split the fraction. Following that, multiplying by the conjugate of the denominator helps convert standard trigonometric fractions into integrable secant/tangent forms.
Step 1: Manipulate the numerator to split the integral.
Add and subtract 1 in the numerator: $$I = \int_{0}^{\pi} \frac{1 + \sin x - 1}{1 + \sin x} dx$$ Split the fraction into two parts: $$I = \int_{0}^{\pi} \left( \frac{1 + \sin x}{1 + \sin x} - \frac{1}{1 + \sin x} \right) dx$$ $$I = \int_{0}^{\pi} \left( 1 - \frac{1}{1 + \sin x} \right) dx$$ $$I = \int_{0}^{\pi} 1 , dx - \int_{0}^{\pi} \frac{1}{1 + \sin x} dx$$ $$I = \pi - \int_{0}^{\pi} \frac{1}{1 + \sin x} dx$$
Step 2: Evaluate the remaining trigonometric integral.
To integrate $\frac{1}{1+\sin x}$, multiply the numerator and denominator by its conjugate $(1-\sin x)$: $$\int_{0}^{\pi} \frac{1}{1+\sin x} \cdot \frac{1-\sin x}{1-\sin x} dx$$ $$= \int_{0}^{\pi} \frac{1-\sin x}{1-\sin^2 x} dx$$ Use the identity $1 - \sin^2 x = \cos^2 x$: $$= \int_{0}^{\pi} \frac{1-\sin x}{\cos^2 x} dx$$
Step 3: Separate and convert to standard integrable forms.
$$= \int_{0}^{\pi} \left( \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} \right) dx$$ $$= \int_{0}^{\pi} (\sec^2 x - \sec x \tan x) dx$$
Step 4: Integrate and apply limits.
The antiderivative is: $$= [\tan x - \sec x]_{0}^{\pi}$$ Evaluate at the upper and lower limits: $$= (\tan \pi - \sec \pi) - (\tan 0 - \sec 0)$$ We know $\tan \pi = 0$, $\sec \pi = -1$, $\tan 0 = 0$, and $\sec 0 = 1$: $$= (0 - (-1)) - (0 - 1)$$ $$= (1) - (-1) = 2$$
Step 5: Calculate final answer.
Substitute this result back into the equation from Step 1: $$I = \pi - 2$$