$\displaystyle\int_0^{\pi/2} \dfrac{\sin x}{\sin x + \cos x}\, dx = ?$
A$0$
B$\pi/4$
C$\pi/2$
D$1$
Answer & Solution
Correct answer: B. $\pi/4$
King's rule (substitute $x \to \pi/2 - x$): $I = \int_0^{\pi/2} \cos x/(\cos x + \sin x)\, dx$.
**Add to original:** $2I = \int_0^{\pi/2} (\sin x + \cos x)/(\sin x + \cos x)\, dx = \pi/2$.
**So $I = \pi/4$.**
Keep practising Unit 6A: Indefinite Integrals & Fundamental Theorem
Browse all Unit 6A: Indefinite Integrals & Fundamental Theorem questions for AP Calculus AB, or explore more Mathematics topics.
Related questions
The integrating factor of the linear differential equation $\dfrac{dy}{dx} + \dfrac{y}{x} The number of arbitrary constants in the general solution of a differential equation of orUsing the substitution $t = 1 + x^2$, $\displaystyle\int \dfrac{x}{1 + x^2}\,dx$ equals:$\displaystyle\int e^x\bigl(f(x) + f'(x)\bigr)\,dx$ equals:$\displaystyle\int \log x\,dx$ equals:$\displaystyle\int ec^2 x\,dx$ equals:$\displaystyle\int \dfrac{1}{ qrt{1 - x^2}}\,dx$ equals:$\displaystyle\int \dfrac{1}{1+x^2}\,dx$ equals: