This page covers how to find indefinite integrals involving trigonometric functions like sine, cosine, tangent, secant, cosecant, and cotangent. These problems use basic trig antiderivative rules, constant multiples, rewriting reciprocal trig functions, and u-substitution. Some challenge problems also use more advanced trig integrals and integration by parts.
Notes
Practice Problems
Find the indefinite integral
\(\textbf{1)}\)\(\displaystyle\int \sin x \,dx \)
The answer is \(-\cos x+C\)
\(\,\,\,\,\,\displaystyle\int \sin x \,dx=-\cos x+C \,\,\,\left(\text{from the notes}\right)\)
\(\textbf{2)}\)\(\displaystyle\int \cos x \,dx \)
The answer is \(\sin x+C\)
\(\,\,\,\,\,\displaystyle\int \cos x \,dx=\sin x+C \,\,\,\left(\text{from the notes}\right)\)
\(\textbf{3)}\)\(\displaystyle\int(5 \sin x-2 \cos x) \,dx \)
The answer is \(-5 \cos x-2 \sin x+C\)
\(\,\,\,\,\,\displaystyle\int(5 \sin x-2 \cos x) \,dx\)
\(\,\,\,\,\,\displaystyle\int 5 \sin x \,dx – \int 2 \cos x \,dx\)
\(\,\,\,\,\,\displaystyle 5 \cdot \int \sin x \,dx – 2 \cdot \int \cos x \,dx\)
\(\,\,\,\,\,\displaystyle 5 \left(-\cos x\right) – 2 \left(\sin x\right)+C\)
\(\,\,\,\,\,\displaystyle -5\cos x – 2\sin x + C\)
\(\textbf{4)}\)\(\displaystyle\int\left(\sec^2 x – \csc^2 x \right) \,dx \)
The answer is \(\tan x+\cot x+C\)
\(\,\,\,\,\,\displaystyle\int\left(\sec^2 x – \csc^2 x \right) \,dx \)
\(\,\,\,\,\,\displaystyle\int\sec^2 x \,dx – \int \csc^2 x \,dx \)
\(\,\,\,\,\,\tan x – \left(-\cot x\right)+C \)
\(\,\,\,\,\,\tan x +\cot x + C \)
\(\textbf{5)}\)\(\displaystyle\int5 \sec x \tan x \,dx \)
The answer is \(5 \sec x+C\)
\(\,\,\,\,\,\displaystyle\int5 \sec x \tan x \,dx \)
\(\,\,\,\,\,\displaystyle 5 \cdot \int\sec x \tan x \,dx \)
\(\,\,\,\,\,5 \sec x+C \)
\(\textbf{6)}\)\(\displaystyle\int -4\sin x \,dx \)
The answer is \(4\cos x+C\)
\(\,\,\,\,\,\displaystyle\int -4\sin x \,dx \)
\(\,\,\,\,\,\displaystyle-4 \cdot \int \sin x \,dx \)
\(\,\,\,\,\,-4 \left(-\cos x\right)+C \)
\(\,\,\,\,\,4\cos x + C \)
\(\textbf{7)}\)\(\displaystyle\int 8 \sec^2 x \,dx \)
The answer is \(8\tan x+C\)
\(\,\,\,\,\,\displaystyle\int 8 \sec^2 x \,dx \)
\(\,\,\,\,\,\displaystyle 8 \cdot \int \sec^2 x \,dx \)
\(\,\,\,\,\,8\tan x+ C \)
\(\textbf{8)}\)\(\displaystyle\int -5\csc x \cot x \,dx \)
The answer is \(5\csc x+C\)
\(\,\,\,\,\,\displaystyle\int -5\csc x \cot x \,dx \)
\(\,\,\,\,\,\displaystyle -5 \cdot \int \csc x \cot x \,dx \)
\(\,\,\,\,\,\displaystyle -5\left(-\csc x\right)+C \)
\(\,\,\,\,\,\displaystyle 5\csc x +C \)
\(\textbf{9)}\)\(\displaystyle\int \frac{5}{ \sec x} \,dx \)
The answer is \(5\sin x+C\)
\(\,\,\,\,\,\displaystyle\int \frac{5}{ \sec x} \,dx \)
\(\,\,\,\,\,\displaystyle\int 5 \cos x \,dx \)
\(\,\,\,\,\,\displaystyle 5 \cdot \int \cos x \,dx \)
\(\,\,\,\,\, 5\sin x + C \)
\(\textbf{10)}\)\(\displaystyle\int \frac{15}{ \csc x} \,dx \)
The answer is \(-15\cos x+C\)
\(\,\,\,\,\,\displaystyle\int \frac{15}{ \csc x} \,dx \)
\(\,\,\,\,\,\displaystyle\int 15\sin x \,dx \)
\(\,\,\,\,\,\displaystyle 15 \cdot \int \sin x \,dx \)
\(\,\,\,\,\, -15\cos x + C \)
\(\textbf{11)}\)\(\displaystyle\int 10 \tan x \,dx \)
The answer is \(-10 \ln |\cos x|+C\)
\(\,\,\,\,\,\displaystyle\int 10 \tan x \,dx \)
\(\,\,\,\,\,\displaystyle 10 \cdot \int \tan x \,dx \)
\(\,\,\,\,\,\displaystyle 10 \cdot \int \frac{\sin x}{\cos x} \,dx \)
\(\,\,\,\,\,u=\cos x, \,\,\, du=-\sin x \,dx \)
\(\,\,\,\,\,\displaystyle 10 \cdot \int -\frac{1}{u} \,du \)
\(\,\,\,\,\,\displaystyle -10 \ln |u| + C \)
\(\,\,\,\,\, -10 \ln \left|\cos x \right|+ C \)
\(\textbf{12)}\)\(\displaystyle\int \cot x \,dx \)
The answer is \(\ln |\sin x|+C\)
\(\,\,\,\,\,\displaystyle\int \cot x \,dx \)
\(\,\,\,\,\,\displaystyle\int \frac{\cos x}{\sin x}\,dx \)
\(\,\,\,\,\,u=\sin x, \,\,\, du=\cos x \,dx \)
\(\,\,\,\,\,\displaystyle\int \frac{1}{u} \,du \)
\(\,\,\,\,\,\ln \left|u\right| + C \)
\(\,\,\,\,\,\ln \left|\sin x\right| + C \)
\(\textbf{13)}\)\(\displaystyle\int 6\cos(3x)\,dx\)
The answer is \(2\sin(3x)+C\)
\(\,\,\,\,\,u=3x\)
\(\,\,\,\,\,du=3\,dx\)
\(\,\,\,\,\,\frac{1}{3}du=dx\)
\(\,\,\,\,\,\displaystyle\int 6\cos(3x)\,dx=\int 6\cos(u)\cdot\frac{1}{3}\,du\)
\(\,\,\,\,\,\displaystyle 2\int \cos(u)\,du\)
\(\,\,\,\,\,2\sin(u)+C\)
\(\,\,\,\,\,2\sin(3x)+C\)
\(\textbf{14)}\)\(\displaystyle\int 4\sin(2x)\,dx\)
The answer is \(-2\cos(2x)+C\)
\(\,\,\,\,\,u=2x\)
\(\,\,\,\,\,du=2\,dx\)
\(\,\,\,\,\,\frac{1}{2}du=dx\)
\(\,\,\,\,\,\displaystyle\int 4\sin(2x)\,dx=\int 4\sin(u)\cdot\frac{1}{2}\,du\)
\(\,\,\,\,\,\displaystyle 2\int \sin(u)\,du\)
\(\,\,\,\,\,-2\cos(u)+C\)
\(\,\,\,\,\,-2\cos(2x)+C\)
\(\textbf{15)}\)\(\displaystyle\int \sin^2 x \,dx\)
The answer is \(\frac{x}{2}-\frac{\sin(2x)}{4}+C\)
\(\,\,\,\,\,\sin^2x=\frac{1-\cos(2x)}{2}\)
\(\,\,\,\,\,\displaystyle\int \sin^2 x\,dx=\int \frac{1-\cos(2x)}{2}\,dx\)
\(\,\,\,\,\,\displaystyle\frac{1}{2}\int 1\,dx-\frac{1}{2}\int \cos(2x)\,dx\)
\(\,\,\,\,\,\displaystyle\frac{x}{2}-\frac{1}{2}\cdot\frac{\sin(2x)}{2}+C\)
\(\,\,\,\,\,\displaystyle\frac{x}{2}-\frac{\sin(2x)}{4}+C\)
\(\textbf{16)}\)\(\displaystyle\int \cos^2 x \,dx\)
The answer is \(\frac{x}{2}+\frac{\sin(2x)}{4}+C\)
\(\,\,\,\,\,\cos^2x=\frac{1+\cos(2x)}{2}\)
\(\,\,\,\,\,\displaystyle\int \cos^2 x\,dx=\int \frac{1+\cos(2x)}{2}\,dx\)
\(\,\,\,\,\,\displaystyle\frac{1}{2}\int 1\,dx+\frac{1}{2}\int \cos(2x)\,dx\)
\(\,\,\,\,\,\displaystyle\frac{x}{2}+\frac{1}{2}\cdot\frac{\sin(2x)}{2}+C\)
\(\,\,\,\,\,\displaystyle\frac{x}{2}+\frac{\sin(2x)}{4}+C\)
\(\textbf{17)}\)\(\displaystyle\int \sin x\cos x \,dx\)
The answer is \(\frac{\sin^2x}{2}+C\)
\(\,\,\,\,\,u=\sin x\)
\(\,\,\,\,\,du=\cos x\,dx\)
\(\,\,\,\,\,\displaystyle\int \sin x\cos x\,dx=\int u\,du\)
\(\,\,\,\,\,\displaystyle\frac{u^2}{2}+C\)
\(\,\,\,\,\,\displaystyle\frac{\sin^2x}{2}+C\)
\(\textbf{18)}\)\(\displaystyle\int \tan^2 x\,dx\)
The answer is \(\tan x-x+C\)
\(\,\,\,\,\,\tan^2x=\sec^2x-1\)
\(\,\,\,\,\,\displaystyle\int \tan^2x\,dx=\int(\sec^2x-1)\,dx\)
\(\,\,\,\,\,\displaystyle\int\sec^2x\,dx-\int1\,dx\)
\(\,\,\,\,\,\tan x-x+C\)
\(\textbf{19)}\)\(\displaystyle\int \sec^2(4x)\,dx\)
The answer is \(\frac{1}{4}\tan(4x)+C\)
\(\,\,\,\,\,u=4x\)
\(\,\,\,\,\,du=4\,dx\)
\(\,\,\,\,\,\frac{1}{4}du=dx\)
\(\,\,\,\,\,\displaystyle\int \sec^2(4x)\,dx=\frac{1}{4}\int\sec^2u\,du\)
\(\,\,\,\,\,\displaystyle\frac{1}{4}\tan u+C\)
\(\,\,\,\,\,\displaystyle\frac{1}{4}\tan(4x)+C\)
\(\textbf{20)}\)\(\displaystyle\int \csc^2(3x)\,dx\)
The answer is \(-\frac{1}{3}\cot(3x)+C\)
\(\,\,\,\,\,u=3x\)
\(\,\,\,\,\,du=3\,dx\)
\(\,\,\,\,\,\frac{1}{3}du=dx\)
\(\,\,\,\,\,\displaystyle\int \csc^2(3x)\,dx=\frac{1}{3}\int\csc^2u\,du\)
\(\,\,\,\,\,-\frac{1}{3}\cot u+C\)
\(\,\,\,\,\,-\frac{1}{3}\cot(3x)+C\)
Challenge Problems
\(\textbf{21)}\)\(\displaystyle\int 5\csc x \,dx \)
The answer is \(-5\ln |\csc x + \cot x| +C\)
\(\,\,\,\,\,\displaystyle\int 5\csc x \,dx \)
\(\,\,\,\,\,\displaystyle5 \cdot \int \csc x \,dx \)
\(\,\,\,\,\,\displaystyle5 \cdot \int \frac{\left(\csc x \right)\left(\csc x + \cot x\right)}{\left(\csc x + \cot x\right)} \,dx \)
\(\,\,\,\,\,\displaystyle5 \cdot \int \frac{\csc^2 x + \cot x \csc x}{\left(\csc x + \cot x\right)} \,dx \)
\(\,\,\,\,\,u=\csc x + \cot x \,\,\, du= -\csc^2x -\cot x \csc x \,dx\)
\(\,\,\,\,\,\displaystyle -5\int \frac{1}{u} \,du \)
\(\,\,\,\,\,-5\ln \left|u\right| + C \)
\(\,\,\,\,\,-5\ln \left|\csc x +\cot x\right| + C \)
\(\textbf{22)}\)\(\displaystyle\int \cos x \sin x \,dx \)
The answer is \(\displaystyle\frac{\sin^2 x}{2}+C\)
\(\,\,\,\,\,\displaystyle\int \cos x \sin x \,dx \)
\(\,\,\,\,\,u=\sin x, \,\,\, du=\cos x \,dx\)
\(\,\,\,\,\,\displaystyle\int u \,du \)
\(\,\,\,\,\,\displaystyle \frac{u^2}{2} +C \)
\(\,\,\,\,\,\displaystyle \frac{\sin^2 x}{2} +C \)
\(\textbf{23)}\)\(\displaystyle\int x \sin x \,dx \)
The answer is \(-x\cos x +\sin x+C\)
\(\,\,\,\,\,\displaystyle\int x \sin x \,dx \)
\(\,\,\,\,\,\displaystyle\int u\,dv=uv-\int v\,du \,\,\, \left(\text{Integration by parts}\right)\)
\(\,\,\,\,\,u=x, \,\,\, dv=\sin x\,dx\)
\(\,\,\,\,\,du=dx, \,\,\, v=-\cos x\)
\(\,\,\,\,\,\displaystyle\int x \sin x \,dx = -x \cos x – \int -\cos x \, dx \)
\(\,\,\,\,\,\displaystyle\int x \sin x \,dx = -x \cos x + \int \cos x \, dx \)
\(\,\,\,\,\,\displaystyle\int x \sin x \,dx = -x \cos x + \sin x + C \)
\(\textbf{24)}\)\(\displaystyle\int -4 \cot x \csc^4 x \,dx \)
The answer is \(\csc^4 x+C\)
\(\,\,\,\,\,\displaystyle\int -4 \cot x \csc^4 x \,dx \)
\(\,\,\,\,\,u=\csc^4x\)
\(\,\,\,\,\,du= 4 \csc^3 x \cdot \left(-\csc x \cot x \right)dx\)
\(\,\,\,\,\,du= -4 \csc^4 x \cot x\,dx\)
\(\,\,\,\,\,\displaystyle\int du \)
\(\,\,\,\,\,u + C \)
\(\,\,\,\,\, \csc^4x + C \)
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