Calculus 2

Evaluate the integral of $\frac{1}{\sqrt{1-x^2}}$ which is equivalent to the inverse sine of x.

For the integral $\displaystyle \sqrt{a^2-x^2}$, make the trigonometric substitution $x=a\sin\theta$ and find the differential $dx=a\cos\theta\,d\theta$.

Evaluate the integral of the form $\displaystyle \int \frac{dx}{\sqrt{x^2 - a^2}}$ by making the substitution $x = a \sec\theta$.

For a radical $\sqrt{x^2 - 1}$, use trigonometric substitution and translate $\sin\theta$ back to $x$ in the problem solved.

Integrate using sine substitution where the substitution is $x = 6 \sin(\theta)$.

Integrate using trig substitution when you have both a radical expression in the numerator and a coefficient on the $x^2$ term.

Using tangent substitution, where $5x=2\tan(\theta)$, solve an integral with a coefficient on $x^2$.

Using secant substitution, where $x = 4 \sec(\theta)$, solve an integral with a rational power, such as a fractional power of three halves.

Evaluate the integral using trigonometric substitution: $\displaystyle \int \sqrt{1-x^2} \, dx$.

Using a secant substitution, simplify and integrate $\frac{1}{x \sqrt{x^2 - 4}}$.

Solve the integral using trigonometric substitution where the square root involves $1 - \sin^2(x)$.

Simplify the integral $\displaystyle \, \int \frac{1}{\sqrt{16 - x^2}} \, dx$ using the substitution $x = 4 \sin \theta$.

Simplify the integral $\displaystyle \int \frac{1}{\sqrt{9 - x^2}} \, dx$ using the substitution $x = 3 \sin \theta$.

Evaluate the integral $\displaystyle \int x^3 \sqrt{1-x^2} \, dx$ using trigonometric substitution.

Evaluate the integral $\displaystyle \int \frac{dx}{\sqrt{x^2 + 16}}$ using trigonometric substitution.

Evaluate the integral $\displaystyle \int \frac{x}{\sqrt{x^2 - 7}} \, dx$ using trigonometric substitution.

Evaluate the indefinite integral $\displaystyle \int \frac{1}{\sqrt{4-x^2}} \, dx$.

Evaluate the integral using trigonometric substitution.

Using the trigonometric substitution $x = a \cdot \sec(\theta)$, simplify the expression involving the square root $\sqrt{x^2 - a^2}$.

Evaluate the integral $\displaystyle \int \frac{x}{\sqrt{25 - x^2}} \, dx$ using trigonometric substitution.