Exercises

Evaluate $\int 4x^3(x^4 + 1)^4 dx$.

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Let

$$u = x^4 + 1$$ $$du = 4x^3$$

That makes the original integral

$$\int 4x^3(x^4 + 1)^4 dx$$ $$\int (x^4 + 1)^4 4x^3 dx$$ $$\int u^4 du$$ $$\frac{1}{5}u^5 + C$$ $$\frac{1}{5}(x^4 + 1)^5 + C$$

Evaluate $\int (x^2 + 2x)(x^3 + 3x^2 + 1)^5 dx$.

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$$\frac{1}{18}(x^3 + 3x^2 + 1)^6 + C$$

Evaluate $\int x^2 \sqrt{x^3 + 2} dx$.

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$$\frac{2}{9}(x^3 + 2)^{3/2} + C = \frac{2}{9}\sqrt{(x^3 + 2)^3} + C$$

Evaluate $\int \frac{x}{\sqrt{3x^2 + 4}} dx$.

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$$\frac{1}{3}\sqrt{3x^2 + 4} + C$$

Evaluate $\int \frac{\sec^2 t}{1 + \tan t} dt$.

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$$\ln|1 + \tan t| + C$$

Evaluate $\int 6^{4x} dx$.

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$$\frac{1}{4\ln 6} 6^{4x} + C$$

Evaluate $\int \frac{x}{x + 3} dx$.

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Let $u = x + 3$ so that $du = dx$. This makes the integral

$$\int \frac{x}{u} du$$

Now, to eliminate the $x$, notice that $x = u - 3$ which makes the integral

$$\int \frac{u - 3}{u} du$$ $$\int\left(1 - \frac{3}{u}\right)du$$ $$u - 3\ln u + C$$ $$x + 3 - 3\ln(x + 3) + C$$

Evaluate $\int \frac{2x^3 + 3x}{x^4 + 3x^2} dx$.

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$$\frac{1}{2}\ln(x^4 + 3x^2) + C$$

Evaluate $\int \left(\cos^2 x\right) dx$.

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Use the substitution $\cos^2 x = \frac{1 + \cos 2x}{2}$.

Evaluate $\int \left(\cos^2 x\right) dx$.

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$$\int \left(\cos^2 x\right) dx$$ $$\int \frac{1 + \cos 2x}{2} dx$$ $$\frac{1}{2}\int\left(1 + \cos 2x\right)dx$$

Now use the substitution $u = 2x$ so $du = 2dx$ or $dx = du/2$.

$$\frac{1}{2}\int\left(1 + \cos u\right)\frac{du}{2}$$ $$\frac{1}{4}\int\left(1 + \cos u\right)du$$ $$\frac{1}{4}\left(u + \sin u\right) + C$$ $$\frac{1}{4}\left(2x + \sin 2x\right) + C$$

Evaluate $\int \frac{\sin x}{\cos^2 x} dx$.

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Method 1: Substitution

Let $u = \cos x$ so $du = -\sin x dx$ or $dx = -\frac{du}{\sin x}$. That makes the integral

$$\int \frac{\sin x}{\cos^2 x} dx$$ $$\int \frac{\sin x}{u^2} \cdot -\frac{du}{\sin x}$$ $$-\int \frac{1}{u^2} du$$ $$-\int u^{-2} du$$ $$-\frac{1}{-1}u^{-1} + C$$ $$(\cos x)^{-1} + C$$ $$\sec x + C$$

Method 2: Trigonometric Simplification

$$\int \frac{1}{\cos x} \cdot {\sin x}{\cos x} dx$$ $$\int \sec x \tan x dx$$ $$\sec x + C$$

Evaluate $\int x\cos(x^2) dx$.

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$$\frac{1}{2}\sin (x^2) + C$$

Evaluate $\int \frac{\cos 2x}{\sin 2x - 1} dx$

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Start with $u = \sin 2x - 1$ so $du = 2\cos 2x dx$ or $dx = \frac{du}{2\cos 2x}$

$$\int \frac{\cos 2x}{\sin 2x - 1} dx$$ $$\int \frac{\cos 2x}{u} \frac{du}{2\cos 2x}$$ $$\frac{1}{2}\int \frac{1}{u} du$$ $$\frac{1}{2}\ln |u| + C$$ $$\frac{1}{2}\ln |\sin 2x - 1| + C$$

$\int\sin^2 x\cos^2 x dx$

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Try using the sine double angle formula, $\sin 2x = 2\sin x\cos x$.

Evaluate $\int\sin^2 x\cos^2 x dx$.

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$$\int\sin^2 x\cos^2 x dx$$ $$\int\left(\sin x\cos x\right)^2 dx$$ $$\int\left(\frac{2\sin x\cos x}{2}\right)^2 dx$$ $$\frac{1}{4}\int (\sin 2x)^2 dx$$

Now substitute $u = 2x$ so $du = 2dx$ or $dx = \frac{du}{2}$.

$$\frac{1}{4}\int (\sin u)^2 \frac{du}{2}$$

Evaluate $\int \cos(x + a) dx$

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$$\int \cos(x + a) = \sin(x + a) + C$$

Notice how the "+ a" doesn't directly impact the result. The antiderivative is shifted by the same amount as the original function.

Evaluate $\int \frac{1}{x + a} dx$

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$$\int \frac{1}{x + a} dx = \ln(x + a) + C$$

The antiderivative here is also just shifted the same amount as the original function.

Suppose $\int f(x) dx = F(x)$. What's $\int f(x + a)dx$?

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Let $u = x + a$ so that $du = dx$.

$$\begin{aligned} \int f(x + a)dx &= \int f(u) du \\ &= F(u) + C \\ &= F(x + a) + C \end{aligned}$$

This formalizes what you saw in the previous three examples: The antiderivative of a translation is the translation of the original function's antiderivative.

Suppose $\int f(x) dx = F(x)$. What's $\int f(nx)dx$?

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Let $u = nx$ so that $du = ndx$ or $dx = \frac{du}{n}$. That makes the integral

$$\begin{aligned} \int f(nx)dx &= \int f(u) \frac{du}{n} \\ &= \frac{1}{n}\int f(u) du \\ &= \frac{1}{n} F(u) + C \\ &= \frac{1}{n} F(nx) + C \end{aligned} $$

Thie result also generalizes the previous three but this time the integral of the scaled function involves both the original scaled variable ($nx$) and a scale factor on the antiderivative ($\frac{1}{n}$).

Show that $\int\left(f(x)\right)^n f'(x) dx = \frac{1}{n+1}\left(f(x)\right)^{n+1} + C, n \ne -1$.

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Let $u = f(x)$ so that $du = f'(x) dx$. This makes the integral

$$\begin{aligned} \int\left(f(x)\right)^n f'(x) dx &= \int u^n du \\ &= \frac{u^{n+1}}{n+1} + C \\ &= \frac{\left(f(x)\right)^{n+1}}{n+1} + C \\ &= \frac{1}{n+1}\left(f(x)\right)^{n+1} + C \end{aligned}$$