MENUFree Step-by-Step Integral Solver
An absolutely free online step-by-step definite and indefinite integrals solver.
Integral Steps:
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Let [tex]u = x^{2}[/tex].
Then let [tex]du = 2 x dx[/tex] and substitute [tex]\frac{du}{2}[/tex]:
[tex]\int \frac{1}{2} \sin{\left (u \right )}\, du[/tex]
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The integral of a constant times a function is the constant times the integral of the function:
[tex]\int \sin{\left (u \right )}\, du = \frac{1}{2} \int \sin{\left (u \right )}\, du[/tex]
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The integral of sine is negative cosine:
[tex]\int \sin{\left (u \right )}\, du = - \cos{\left (u \right )}[/tex]
So, the result is: [tex]- \frac{1}{2} \cos{\left (u \right )}[/tex]
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Now substitute [tex]u[/tex] back in:
[tex]- \frac{1}{2} \cos{\left (x^{2} \right )}[/tex]
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Add the constant of integration:
[tex]- \frac{1}{2} \cos{\left (x^{2} \right )}+ \mathrm{constant}[/tex]
The answer is:
[tex]- \frac{1}{2} \cos{\left (x^{2} \right )}+ \mathrm{constant}[/tex]
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* is multiplication
oo is $\infty$
pi is $\pi$
x^2 is x2
sqrt(x) is $\sqrt{x}$
sqrt[3](x) is $\sqrt[3]{x}$
(a+b)/(c+d) is $\frac{a+b}{c+d}$
Common Integrals
$\int 0dx = \text{const}$
$\int\ dx = x + \text{const}$
$\int kdx = kx + \text{const}$
$\int x^n\ dx = \frac{1}{n+1}x^{n+1} + \text{const}$ here n≠-1
$\int \frac{1}{x}\ dx = \int x^{-1}\ dx = \ln|x| + \text{const}$
$\int x^{-n}\ dx = \frac{1}{-n+1}x^{-n+1} + \text{const}$
$\int \frac{1}{ax+b}\ dx = \frac{1}{a}\ln|ax+b| + \text{const}$
$\int e^x\ dx = e^x + \text{const}$
$\int a^x\ dx = \frac{a^x}{\\ln a} + \text{const}$
$\int \sin(x)\ dx = -\cos(x) + \text{const}$
$\int \cos(x)\ dx = \sin(x) + \text{const}$
$\int \tan(x)\ dx = \ln|sec(x)| + \text{const}$
$\int \cot(x)\ dx = \ln|\sin(x)| + \text{const}$
$\int \frac{1}{\sqrt{1-x^2}} \ dx = \arcsin(x) + \text{const}$
$\int -\frac{1}{\sqrt{1-x^2}} \ dx = \arccos(x) + \text{const}$
$\int \frac{1}{1+ x^2}\ dx = \arctan(x) + \text{const}$
$\int -\frac{1}{1+x^2}\ dx = \text{arccot}(x) + \text{const}$
Integration by Parts
$\int u\ dv = uv - \int v\ du$
$\int\limits_{a}^{b} u\ dv = uv |_a^b - \int v\ du$
Trigonometric Substitutions
$\sqrt{a^2 - b^2x^2}$ $\Rightarrow x=\frac{a}{b}\sin\theta$ and $\cos^2\theta = 1 - \sin^2\theta$
$\sqrt{a^2 + b^2x^2}$ $\Rightarrow x=\frac{a}{b}\tan\theta$ and $\sec^2\theta = 1 + \tan^2\theta$
$\sqrt{b^2x^2 - a^2}$ $\Rightarrow x=\frac{a}{b}\sec\theta$ and $\tan^2\theta = \sec^2\theta - 1$
