Integrate: $\int \frac{1}{\sin x+\sin 2x}\,dx$

Integrate: ∫ (1)/(sin x+sin 2x)dx

Miscellaneous 3 · Q215

\int \frac{1}{\sin x+\sin 2x}\,dx

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★

17% · 44/255 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

$\sin x+\sin2x=\sin x+2\sin x\cos x=\sin x(1+2\cos x)$ . Multiply numerator and denominator of $\frac{1}{\sin x(1+2\cos x)}$ by $\sin x$ : $\frac{\sin x}{\sin^2x(1+2\cos x)}=\frac{\sin x}{(1-\cos^2x)(1+2\cos x)}$ . Let $u=\cos x$ , $du=-\sin x\,dx$ : the integral becomes $-\int\frac{du}{(1-u)(1+u)(1+2u)}$ . Partial fractions: $\frac{1}{(1-u)(1+u)(1+2u)}=\frac{1/6}{1-u}-\frac{1/2}{1+u}+\frac{4/3}{1+2u}$ (found by evaluating at $u=1,-1,-\frac12$ , checked at $u=0$ : $\frac16-\frac12+\frac43=1$ ✓). Integrating with the leading minus sign: $-\int\left[\frac{1/6}{1-u}-\frac{1/2}{1+u}+\frac{4/3}{1+2u}\right]du=\frac16\log|1-u|+\frac12\log|1+u|-\frac23\log|1+2u|+c$ …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 mark
$\int \frac{x}{(1 + x)(2 + x)}\,dx = \int \left(\frac{A}{1 + x} + \frac{B}{2 + x}\right)dx$
›Reveal solutionSolution

Partial fractions give $A = -1,\; B = 2$ , so the claim $A = 1,\, B = 2$ is False.

Decompose the integrand:

$\frac{x}{(1 + x)(2 + x)} = \frac{A}{1 + x} + \frac{B}{2 + x}.$

Multiply through by $(1+x)(2+x)$ :

$x = A(2 + x) + B(1 + x).$

Put $x = -1$ : $\;-1 = A(2 - 1) + B(0) = A \Rightarrow A = -1.$

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$\displaystyle\int \dfrac{dx}{(x - 8)(x + 7)} =$
(a) $\dfrac{1}{15} \log \left|\dfrac{x + 2}{x - 1}\right| + c$
(b) $\dfrac{1}{15} \log \left|\dfrac{x + 8}{x + 7}\right| + c$
(c) $\dfrac{1}{15} \log \left|\dfrac{x - 8}{x + 7}\right| + c$
(d) $(x - 8)(x - 7) + c$
(e) $\dfrac{1}{15} \log \left|\dfrac{x + 2}{x + 1}\right| + c$
(f) $(x - 8)(x + 7) + c$
›Reveal solutionSolution

Partial fractions give $\dfrac{1}{(x-8)(x+7)} = \dfrac{1}{15}\!\left(\dfrac{1}{x-8} - \dfrac{1}{x+7}\right)$ , which integrates to $\dfrac{1}{15}\log\left|\dfrac{x-8}{x+7}\right| + c$ .

Write

$\frac{1}{(x-8)(x+7)} = \frac{A}{x-8} + \frac{B}{x+7} \;\Rightarrow\; A(x+7) + B(x-8) = 1.$

Put $x = 8$ : $\;A(15) = 1 \Rightarrow A = \dfrac{1}{15}$ .

Put $x = -7$ : $\;B(-15) = 1 \Rightarrow B = -\dfrac{1}{15}$ .

Hence …

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 mark
$\int \dfrac{x}{(x + 2)(x + 3)}\, dx =$
›Reveal solutionSolution
Partial fractions give $\dfrac{x}{(x+2)(x+3)} = \dfrac{-2}{x+2} + \dfrac{3}{x+3}$ , so the blank preceding $\displaystyle\int \dfrac{3}{x+3}\,dx$ is $\displaystyle\int \dfrac{-2}{x+2}\,dx$ .

Resolve the integrand into partial fractions:

$\frac{x}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3}.$

Clearing denominators: $x = A(x+3) + B(x+2)$ .

Put $x = -2$ : $-2 = A(1) \Rightarrow A = -2$ .

Put $x = -3$ : $-3 = B(-1) \Rightarrow B = 3$ .

Hence

$\frac{x}{(x+2)(x+3)} = \frac{-2}{x+2} + \frac{3}{x+3},$

so …
Start your 14-day free trial to see the full solution →

←Previous questionQ214 — Integrate: ∫ (x²)/((x-1)(3x-1)(3x-2))dx →Next questionQ216 — Integrate: ∫ sec² x√(7+2tan x-tan² x)dx

Q.Integrate: $\int \frac{1}{\sin x+\sin 2x}\,dx$

Concept understanding — Integration by Partial Fractions

Q.Integrate: ∫1sin⁡x+sin⁡2x dx\int \frac{1}{\sin x+\sin 2x}\,dx∫sinx+sin2x1​dx

🎓Unlock everything free for 14 days

Q.Integrate: $\int \frac{1}{\sin x+\sin 2x}\,dx$

Unlock everything free for 14 days