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Calculus 2

Simplify and integrate 4x22x8dx\displaystyle \int \frac{4}{x^2 - 2x - 8} \, dx by completing the square.

Solve the indefinite integral 4x216x2dx\displaystyle \int 4x^2 \sqrt{16 - x^2} \, dx using appropriate substitution.

Indefinite integral of 1x225dx\displaystyle \int \frac{1}{\sqrt{x^2 - 25}} \, dx using trigonometric substitution.

Integrate the integral of x31x2\frac{x^3}{\sqrt{1-x^2}} with respect to xx.

Evaluate the integral of 25x24x\displaystyle \frac{\sqrt{25x^2 - 4}}{x} with respect to xx.

Evaluate the definite integral 03xx2+9dx\displaystyle \int_{0}^{3} x \sqrt{x^2 + 9} \, dx.

Find the area bounded by the ellipse 2x2+9y2=12x^2 + 9y^2 = 1.

Solve the integral of x+23(x2+4x)(x+2)3/2dx\displaystyle \int \frac{\sqrt{x+2}^3(x^2+4x)}{(x+2)^{-3/2}} \, dx by completing the square and using a trigonometric substitution.

Given the integral with a radical, use the substitution x=3sinθx = 3\sin\theta to simplify 9x2\sqrt{9 - x^2} and find the integral.

Using trigonometric identities, such as cos2(θ)+sin2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1, find related identities to simplify expressions in integral problems.

Suppose 4x2+1=sec2(θ)4x^2 + 1 = \sec^2(\theta). Use the substitution x=12tan(θ)x = \frac{1}{2} \tan(\theta) to find the integral.

Given the integral exe2x9e2xdx\displaystyle \int \frac{e^x}{e^{2x} \sqrt{9 - e^{2x}}} \, dx, use trigonometric substitution to evaluate it.

Evaluate the integral dx(x3)29(x3)2 \displaystyle \int \frac{dx}{(x-3)^2 \sqrt{9 - (x-3)^2}} .

Make the substitution u=exu = e^x to evaluate the integral duu29u2 \int \frac{du}{u^2 \sqrt{9 - u^2}}.

Using the substitution x=3sec(θ)x = \sqrt{3} \sec(\theta), evaluate the integral x23xdx\displaystyle \int \frac{\sqrt{x^2 - 3}}{x} \, dx.

Calculate the definite integral of sec2(θ)1\sec^2(\theta) - 1 from θ=0\theta = 0 to θ=π6\theta = \frac{\pi}{6}.

Integrate cos3θ\cos^3 \theta with respect to θ\theta.

Evaluate the integral 9x2x2dx\displaystyle \int \frac{\sqrt{9-x^2}}{x^2} \, dx using trigonometric substitution.

Evaluate the integral 1x2x2+4dx\displaystyle \int \frac{1}{x^2 \sqrt{x^2+4}} \, dx using trigonometric substitution.