Trigonometric Integrals and Substitution
Two related but distinct tools live here. Trig integrals use identities to simplify powers of sin/cos before integrating: for odd powers, save one factor and use Pythagorean substitution ($\sin^2x=1-\cos^2x$); for even powers, use power-reduction ($\sin^2x=\tfrac{1-\cos2x}{2}$) to lower the degree. Trig substitution is a technique to eliminate square roots of the form $\sqrt{a^2\pm x^2}$ or $\sqrt{x^2-a^2}$ by swapping in a trig function whose Pythagorean identity collapses the radical. After integrating in $\theta$, always draw a right triangle from the original substitution to convert back to $x$. If the expression isn't in standard form yet, complete the square first.
Form under radical Substitution Simplification
$\sqrt{a^2-x^2}$ $x=a\sin\theta$ $\sqrt{a^2-x^2}=a\cos\theta$ $\sqrt{a^2+x^2}$ $x=a\tan\theta$ $\sqrt{a^2+x^2}=a\sec\theta$ $\sqrt{x^2-a^2}$ $x=a\sec\theta$ $\sqrt{x^2-a^2}=a\tan\theta$
Product-to-Sum Identities
$$2\sin A\sin B=\cos(A-B)-\cos(A+B)$$
$$2\cos A\cos B=\cos(A-B)+\cos(A+B)$$
$$2\sin A\cos B=\sin(A-B)+\sin(A+B)$$
Power-Reduction and Pythagorean Identities
$$\sin^2 x=\frac{1-\cos 2x}{2}$$
$$\cos^2 x=\frac{1+\cos 2x}{2}$$
$$1+\tan^2 x=\sec^2 x$$
$$1+\cot^2 x=\csc^2 x$$
Basic Trigonometric Integrals
$$\int \sin u\,du = -\cos u + C$$
$$\int \cos u\,du = \sin u + C$$
$$\int \sec^2 u\,du = \tan u + C$$
$$\int \csc^2 u\,du = -\cot u + C$$
$$\int \sec u\,\tan u\,du = \sec u + C$$
$$\int \csc u\,\cot u\,du = -\csc u + C$$
$$\int \tan u\,du = \ln|\sec u| + C$$
$$\int \cot u\,du = \ln|\sin u| + C$$
$$\int \sec u\,du = \ln|\sec u+\tan u| + C$$
$$\int \csc u\,du = \ln|\csc u-\cot u| + C$$
$$\int \frac{du}{a^2+u^2}=\frac{1}{a}\tan^{-1}\!\left(\frac{u}{a}\right)+C$$
$$\int \frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\!\left(\frac{u}{a}\right)+C$$
$$\int \frac{du}{a^2-u^2}=\frac{1}{2a}\ln\left|\frac{a+u}{a-u}\right|+C$$
$$\int \frac{du}{\sqrt{u^2-a^2}}=\ln\left|u+\sqrt{u^2-a^2}\right|+C$$
Trigonometric Substitution
$$x = a\sin\theta$$
$$dx = a\cos\theta\,d\theta$$
$$\sqrt{a^2 - x^2} = a\cos\theta$$
$$x = a\tan\theta$$
$$dx = a\sec^2\theta\,d\theta$$
$$\sqrt{a^2 + x^2} = a\sec\theta$$
$$x = a\sec\theta$$
$$dx = a\sec\theta\tan\theta\,d\theta$$
$$\sqrt{x^2 - a^2} = a\tan\theta$$
Back-Substitution Using Right Triangles
$$\sin\theta=\frac{x}{a}\qquad \cos\theta=\frac{\sqrt{a^2-x^2}}{a}$$
$$\tan\theta=\frac{x}{a}\qquad \sec\theta=\frac{\sqrt{a^2+x^2}}{a}$$
$$\sec\theta=\frac{x}{a}\qquad \tan\theta=\frac{\sqrt{x^2-a^2}}{a}$$
Evaluate $$\int_0^{\pi/2}2\sin^6x\cos^4x\,dx.$$
Show Solution
Using the beta/Wallis result,
$$\int_0^{\pi/2}\sin^6x\cos^4x\,dx=\frac{1}{2}B\left(\frac{7}{2},\frac{5}{2}\right)=\frac{3\pi}{512}$$
Therefore,
$$\int_0^{\pi/2}2\sin^6x\cos^4x\,dx=\frac{3\pi}{256}=\boxed{0.0368}$$
Even-Power Cosine Integral
Evaluate $$\int_0^{\pi/2}\cos^8x\,dx.$$
Show Solution
For an even power,
$$\int_0^{\pi/2}\cos^8x\,dx=\frac{7\cdot5\cdot3\cdot1}{8\cdot6\cdot4\cdot2}\cdot\frac{\pi}{2}$$
$$=\frac{35\pi}{256}=\boxed{0.4295}$$
Area Between Sine and Cosine on a Short Interval
Find the area between $y=\sin x$ and $y=\cos x$ from $x=0$ to $x=\pi/4$.
Show Solution
On $[0,\pi/4]$, $\cos x$ is above $\sin x$.
$$A=\int_0^{\pi/4}(\cos x-\sin x)\,dx=\left[\sin x+\cos x\right]_0^{\pi/4}$$
$$A=\sqrt{2}-1=\boxed{0.414}$$
First-Quadrant Area With Sine, Cosine, and Axis
Find the area bounded by $y=\sin x$, $y=\cos x$, and the $x$-axis in the first quadrant.
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The curves meet at $x=\pi/4$. The region equals the area under $y=\cos x$ from $0$ to $\pi/2$ minus the area between $\cos x$ and $\sin x$ from $0$ to $\pi/4$.
$$A=\int_0^{\pi/2}\cos x\,dx-(\sqrt{2}-1)=1-\sqrt{2}+1$$
$$A=2-\sqrt{2}=\boxed{0.586\text{ sq. units}}$$
Evaluate $$\int\frac{dx}{\sqrt{25-x^2}}.$$
Show Solution
Use $x=5\sin\theta$, so $dx=5\cos\theta\,d\theta$ and $\sqrt{25-x^2}=5\cos\theta$.
$$\int\frac{dx}{\sqrt{25-x^2}}=\int d\theta=\theta+C$$
Since $\theta=\sin^{-1}(x/5)$,
$$\boxed{\sin^{-1}\left(\frac{x}{5}\right)+C}$$
Case 1 Full Back-Substitution: $\sqrt{a^2-x^2}$
Evaluate $$\int x^2\sqrt{4-x^2}\,dx.$$
Show Solution
Let $x=2\sin\theta$, $dx=2\cos\theta\,d\theta$, $\sqrt{4-x^2}=2\cos\theta$.
$$\int x^2\sqrt{4-x^2}\,dx=\int(2\sin\theta)^2(2\cos\theta)(2\cos\theta\,d\theta)=16\int\sin^2\theta\cos^2\theta\,d\theta$$
Use $\sin^2\theta\cos^2\theta=\frac{1}{4}\sin^2(2\theta)=\frac{1-\cos4\theta}{8}$:
$$=16\int\frac{1-\cos4\theta}{8}\,d\theta=2\left(\theta-\frac{\sin4\theta}{4}\right)+C$$
Back-substitute using the triangle from $x=2\sin\theta$: $\sin\theta=x/2$, $\cos\theta=\sqrt{4-x^2}/2$, so $\sin2\theta=2\sin\theta\cos\theta=\frac{x\sqrt{4-x^2}}{2}$ and $\sin4\theta=2\sin2\theta\cos2\theta$. After simplification:
$$\boxed{\frac{x(2x^2-4)\sqrt{4-x^2}}{8}+\frac{1}{2}\sin^{-1}\!\left(\frac{x}{2}\right)+C}$$
Case 2 Full Example: $\sqrt{a^2+x^2}$
Evaluate $$\int\frac{dx}{(x^2+9)^{3/2}}.$$
Show Solution
Let $x=3\tan\theta$, $dx=3\sec^2\theta\,d\theta$, $(x^2+9)^{3/2}=(9\sec^2\theta)^{3/2}=27\sec^3\theta$.
$$\int\frac{dx}{(x^2+9)^{3/2}}=\int\frac{3\sec^2\theta\,d\theta}{27\sec^3\theta}=\frac{1}{9}\int\cos\theta\,d\theta=\frac{\sin\theta}{9}+C$$
Back-substitute : from $x=3\tan\theta$, draw a right triangle with opposite $=x$, adjacent $=3$, hypotenuse $=\sqrt{x^2+9}$. So $\sin\theta=\frac{x}{\sqrt{x^2+9}}$.
$$\boxed{\frac{x}{9\sqrt{x^2+9}}+C}$$
Case 3 Full Example: $\sqrt{x^2-a^2}$
Evaluate $$\int\frac{\sqrt{x^2-4}}{x}\,dx.$$
Show Solution
Let $x=2\sec\theta$, $dx=2\sec\theta\tan\theta\,d\theta$, $\sqrt{x^2-4}=2\tan\theta$.
$$\int\frac{2\tan\theta}{2\sec\theta}\cdot2\sec\theta\tan\theta\,d\theta=2\int\tan^2\theta\,d\theta=2\int(\sec^2\theta-1)\,d\theta$$
$$=2(\tan\theta-\theta)+C$$
Back-substitute : $\sec\theta=x/2$, so $\tan\theta=\frac{\sqrt{x^2-4}}{2}$ and $\theta=\sec^{-1}(x/2)$.
$$\boxed{\sqrt{x^2-4}-2\sec^{-1}\!\left(\frac{x}{2}\right)+C}$$
Completing the Square Before Trig Sub
Evaluate $$\int\frac{dx}{\sqrt{-x^2+4x}}.$$
Show Solution
Step 1 — Complete the square in the radicand: $-x^2+4x=-(x^2-4x)=-(x^2-4x+4-4)=4-(x-2)^2$.
$$\int\frac{dx}{\sqrt{4-(x-2)^2}}$$
Step 2 : Let $u=x-2$, so $du=dx$:
$$\int\frac{du}{\sqrt{4-u^2}}=\sin^{-1}\!\left(\frac{u}{2}\right)+C$$
Back-substitute $u=x-2$:
$$\boxed{\sin^{-1}\!\left(\frac{x-2}{2}\right)+C}$$
Definite Integral with Trig Substitution
Evaluate $$\int_0^{2}\frac{x^3}{\sqrt{4-x^2}}\,dx.$$
Show Solution
Let $x=2\sin\theta$, $dx=2\cos\theta\,d\theta$, $\sqrt{4-x^2}=2\cos\theta$. Change limits: $x=0\Rightarrow\theta=0$; $x=2\Rightarrow\theta=\pi/2$.
$$\int_0^{\pi/2}\frac{8\sin^3\theta}{2\cos\theta}\cdot2\cos\theta\,d\theta=8\int_0^{\pi/2}\sin^3\theta\,d\theta$$
Split $\sin^3\theta=\sin\theta(1-\cos^2\theta)$ and let $u=\cos\theta$:
$$=8\int_1^0-(1-u^2)\,du=8\int_0^1(1-u^2)\,du=8\left[u-\frac{u^3}{3}\right]_0^1=8\cdot\frac{2}{3}$$
$$\boxed{\frac{16}{3}}$$
Arc Length Using Trig Sub
Find the arc length of the parabola $y=\tfrac{x^2}{2}$ from $x=0$ to $x=1$.
Show Solution
Arc length formula: $L=\int_0^1\sqrt{1+(y')^2}\,dx$. Since $y'=x$:
$$L=\int_0^1\sqrt{1+x^2}\,dx$$
Let $x=\tan\theta$, $dx=\sec^2\theta\,d\theta$, $\sqrt{1+x^2}=\sec\theta$. Limits: $\theta=0$ to $\theta=\pi/4$.
$$L=\int_0^{\pi/4}\sec^3\theta\,d\theta=\left[\frac{\sec\theta\tan\theta}{2}+\frac{\ln|\sec\theta+\tan\theta|}{2}\right]_0^{\pi/4}$$
$$=\frac{\sqrt{2}\cdot1}{2}+\frac{\ln(\sqrt{2}+1)}{2}=\frac{\sqrt{2}}{2}+\frac{\ln(\sqrt{2}+1)}{2}$$
$$\boxed{L\approx1.1478}$$
