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Integration of Trigonometric Functions - Practice Questions & MCQ

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

Quick Facts

  • Fundamental Formulae of Indefinite Integration (Trigonometric Functions) is considered one of the most asked concept.

  • 22 Questions around this concept.

Solve by difficulty

The integral \small \int \sqrt{1 + 2cot x ( cosec x + cot x) dx} \small \left ( 0< x< \frac{\pi }{2} \right ) is equal to (where C is a constant of integration)

Evaluate the integral of \int\left ( 4sin\ x+3cos\ x-2sec^{2}x \right )dx.

Concepts Covered - 1

Fundamental Formulae of Indefinite Integration (Trigonometric Functions)

Trigonometric Functions

\\\mathrm{1.\;\;\frac{d}{d x}(-\cos x)=\sin x \Rightarrow \int \sin x\; d x=-\cos x+C}\\\\\mathrm{2.\;\;\frac{d}{d x}(sin x)=\cos x \Rightarrow \int \cos x\; d x=\sin x+C}\\\\\mathrm{3.\;\;\frac{d}{d x}(\tan x)=\sec ^{2} x \Rightarrow \int \sec ^{2} x \;d x=\tan x+C}\\\\\mathrm{4.\;\;\frac{d}{d x}(-\cot x)=\csc ^{2} x \Rightarrow \int \csc ^{2} x\; d x=-\cot x+C}\\\\\mathrm{5.\;\;\frac{d}{d x}(\sec x)=\sec x \tan x\;\Rightarrow \int \sec x \tan x\; d x=\sec x+C}\\\\\mathrm{6.\;\;{\frac{d}{d x}(-\csc x)=\csc x \cot x} \\ {\;\Rightarrow \int \csc x \cot x \;d x=-\csc x+C}}

Integrals of tan x, cot x, sec x, cosec x

\\\\\mathrm{7.\;\;{\frac{d}{d x}(\log |\sin x|)=\cot x} \\ {\;\Rightarrow \int \cot x \;d x=\log |\sin x|+C}}\\\\\mathrm{8.\;\;{\frac{d}{d x}(-\log |\cos x|)=\tan x} \\ {\;\Rightarrow \int \tan x \;d x=-\log |\cos x|+C}}\\\\\mathrm{9.\;\;{\frac{d}{d x}(\log |\sec x+\tan x|)=\sec x} \\ {\;\Rightarrow \int \sec x \;d x=\log |\sec x+\tan x|+C}}\\\\\mathrm{10.\;\;{\frac{d}{d x}(\log |\csc x-\cot x|)=\csc x} \\ {\;\Rightarrow \int \csc x \;d x=\log |\csc x-\cot x|+C}}

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Fundamental Formulae of Indefinite Integration (Trigonometric Functions)

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Fundamental Formulae of Indefinite Integration (Trigonometric Functions)

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 7.1

Line : 40

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