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Trigonometric Integrals - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

15 Questions around this concept.

Solve by difficulty

If $\int \frac{2\cos x-\sin x+\lambda }{\cos x+\sin x-2}dx=A\: ln\left | \cos x+\sin x-2 \right |+Bx+C$ Then the ordered triplet $A,B,\lambda$ is

$\left ( \frac{1}{1},\frac{3}{2},-1 \right )$

$\left ( \frac{3}{2},\frac{1}{2},-1 \right )$

$\left ( \frac{1}{1},-1,-\frac{3}{2} \right )$

$\left ( \frac{3}{2},-1,\frac{1}{2} \right )$

View Solution

Concepts Covered - 2

Trigonometric Integrals (Part 1)

(a) Integral of the form

$\\\mathrm{1.\;\;\int \frac{1}{a \cos ^{2} x+b \sin ^{2} x} \;d x\;\;\;\;\;\;}\\\\\mathrm{2.\;\;\int \frac{1}{a +b \sin ^{2} x} \;d x}\\\\\mathrm{3.\;\;\int \frac{1}{a+b \cos ^{2} x} \;d x\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\\\\mathrm{4.\;\;\int \frac{1}{a+b \sin ^{2} x+c \cos ^{2} x} \;d x}$

Working Rule:

Step 1 : Divide numerator and denominator both by cos²x.

Step 2 : Put tan x = t, sec²x dx = dt

This substitution will convert the trigonometric integral into algebraic integral.

After employing these steps the integral will reduce to the form $\int \frac{f(t) d t}{A t^{2}+Bt+C}$ , where f(t) is a polynomial in t.

This integral can be evaluated by methods we studied in previous concepts.

(b) Integral of the form

$\\\mathrm{1.\;\;\int \frac{1}{a \sin x+b \cos x} \;d x}\\\\\mathrm{2.\;\;\int \frac{1}{a+b \sin x} \;d x}\\\\\mathrm{3.\;\;\int \frac{1}{a+b \cos x} \;d x, }\\\\\mathrm{4.\;\;\int \frac{1}{a \sin x+b \cos x+c}\; d x}$

Working Rule:

Write sin x and cos x in terms of tan (x/2) and then substitute for tan (x/2) = t

i.e.

$\\ \sin x=\frac{2 \tan x / 2}{1+\tan ^{2} x / 2} \text { and } \cos x=\frac{1-\tan ^{2} x / 2}{1+\tan ^{2} x / 2}\\\\\mathrm{replace,\;\;\tan(x/2)\;with\;\mathit{t}}\\\text {by performing these steps the integral reduces to the form}\\\\\mathrm{\int \frac{1}{a t^{2}+b t+c} d t\;\;which \;can\;be\;solved\;by\;method\;we\;studied\;in}\\\mathrm{previous\;concepts.}$

Trigonometric Integrals (Part 2)

(c) Integrals of the form

$\\\mathrm{1.\;\;\int \frac{p \cos x+q \sin x+r}{a \cos x+b \sin x+c} \;d x}\\\\\\\mathrm{2.\;\;\int \frac{p \cos x+q \sin x}{a \cos x+b \sin x}\,\, d x}$

Working Rule:

$\\\mathrm{Express \;numerator\;as\;{\lambda(\text { denominator })+\mu(\text { differentiation of denominator })+\gamma}}$

$\\\Rightarrow \mathrm{\left (p\cos x+q\sin x+r \right )=\lambda\left ( a\cos x+b\sin x+c \right )+\mu\left ( -a\sin x+b\cos x \right )+\gamma}$

where λ, μ and γ are constants to be determined by comparing the coefficients of sinx, cos x and constant
terms on both sides.

$\\\mathrm{\int \frac{p \cos x+q \sin x+r}{a \cos x+b \sin x+c} \;d x=\int \frac{\lambda\left ( a\cos x+b\sin x+c \right )+\mu\left ( -a\sin x+b\cos x \right )+\gamma}{a\cos x+b\sin x+c}dx}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\lambda\int\frac{a\cos x+b\sin x+c}{a\cos x+b\sin x+c}dx+\mu\int\frac{-a\sin x+b\cos x}{a\cos x+b\sin x+c}dx\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\int \frac{\mu}{a\cos x+b\sin x+c}dx}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\lambda x+\mu\ln\left | a\cos x+b\sin x+c \right |+\int \frac{\mu}{a\cos x+b\sin x+c}dx$