Integrals - GRE Subject Test: Math

Card 0 of 14

Question

Evaluate the following integral.

$\int x e^{4x}dx$

Answer

Integration by parts follows the formula:

$\int u \ dv=uv-\int v\ du$

In this problem we have $\int x e^{4x}dx$ so we'll assign our substitutions:

and $dv=e^{4x} dx$

which means and $v=\frac{1}{4}e^{4x}$

Including our substitutions into the formula gives us:

$x\frac{1}{4}e^{4x}-\int \frac{1}{4}e^{4x}dx$

We can pull out the fraction from the integral in the second part:

$\frac{x}{4}e^{4x}-\frac{1}{4} \int e^{4x}dx$

Completing the integration gives us:

$\frac{x}{4}e^{4x}-\frac{1}{4} \left(\frac{1}{4}e^{4x}\right) +c$

$\frac{x}{4}e^{4x}-\frac{1}{16} e^{4x} +c$

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Question

Integrate the following.

$\int x \cos x \ dx$

Answer

Integration by parts follows the formula:

$\int u \ dv=uv- \int v \ du$

So, our substitutions will be and $dv=\cos x \ dx$

which means and $v=\sin x$

Plugging our substitutions into the formula gives us:

$x\sin x-\int \sin x \ dx$

Since $\int \sin x \ dx=-\cos x + c$ , we have:

$x\sin x - (- \cos x + c)$ , or

$x\sin x + \cos x + c$

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Question

Evaluate the following integral.

$\int x^2 \ln x \ dx$

Answer

Integration by parts follows the formula:

$\int u \ dv=uv-\int v \ du$

Our substitutions will be $u=\ln x$ and

which means $du= \frac{1}{x}dx$ and $v=\frac{1}{3}x^3$ .

Plugging our substitutions into the formula gives us:

$\frac{x^3}{3} \ln x -\int \frac{x^3}{3} \cdot \frac{1}{x} dx$

Look at the integral: we can pull out the $\frac{1}{3}$ and simplify the remaining $\frac{x^3}{x}$ as

$\frac{x^3}{3} \ln x -\frac{1}{3} \int x^2 dx$ .

We now solve the integral: $\int x^2 dx = \frac{1}{3} x^3 + c$ , so:

$\frac{x^3}{3} \ln x -\frac{1}{3} \left(\frac{1}{3}x^3 + c\right)$

$\frac{x^3}{3} \ln x - \frac{1}{9} x^3+c$

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Question

Evaluate the following integral.

$\int x \cos (2x+1)dx$

Answer

Integration by parts follows the formula:

$\int u \ dv=uv - \int v \ du$ .

Our substitutions are and $dv= \cos (2x+1) dx$

which means and $v=\frac{1}{2} \sin (2x+1)$ .

Plugging in our substitutions into the formula gives us

$\frac{x}{2} \sin (2x+1)-\int \frac{1}{2} \sin (2x+1) dx$

We can pull $\frac{1}{2}$ outside of the integral.

$\frac{x}{2} \sin (2x+1)-\frac{1}{2} \int \sin (2x+1) dx$

Since $\int \sin (2x+1) dx =-\frac{1}{2} \cos (2x+1)+ c$ , we have

$\frac{x}{2} \sin (2x+1)- \frac{1}{2} \left(-\frac{1}{2} \cos (2x+1) + c\right)$

$\frac{x}{2} \sin (2x+1)+\frac{1}{4} \cos (2x+1) + c$

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Question

Integrate:

$\int \frac{18x-6}{3x^2-2x}dx$

Answer

This problem requires U-Substitution. Let and find .

Notice that the numerator in $\int \frac{18x-6}{3x^2-2x}dx$ has common factor of 2, 3, or 6. The numerator can be factored so that the term can be a substitute. Factor the numerator using 3 as the common factor.

$\int \frac{18x-6}{3x^2-2x}dx= \int\frac{3(6x-2)}{3x^2-2x}dx=3\int \frac{(6x-2)}{3x^2-2x}dx$

Substitute and terms, integrate, and resubstitute the term.

$3\int \frac{du}{u} = 3: ln\left | u\right |+C=3: ln\left | 3x^2-2x\right |+C$

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Question

Evaluate the following integral:

$\int5x^4(x^5-9)^6dx$

Answer

To calculate this integral, we could expand that whole binomial, but it would be very time consuming and a bit of a pain. Instead, let's use u substitution:

Given this:

$\int5x^4(x^5-9)^6dx$

We can say that

Then, plug it back into our original expression

$\int u^6du$

Evaluate this integral to get

$\frac{u^7}{7}+c$

Then, replace u with what we substituted it for to get our final answer. Note because this is an indefinite integral, we need a plus c in it.

$\frac{(x^5-9)^7}{7}+c$

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Question

Integrate the following using substitution.

$\int \frac{1}{4-5x}dx$

Answer

$\int \frac{1}{4-5x}dx$

Using substitution, we set which means its derivative is .

Substituting for , and $- \frac{1}{5}du$ for we have:

$\int \frac{1}{4-5x}dx=-\frac{1}{5} \int \frac{1}{u} du$

Now we just integrate:

$-\frac{1}{5} \int \frac{1}{u} du=-\frac{1}{5}\ln\left | u\right |+c$

Finally, we remove our substitution to arrive at an expression with our original variable:

$-\frac{1}{5}\ln\left | 4-5x\right |+c$

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Question

Evaluate the following integral:

$\int5x^4(x^5-9)^6dx$

Answer

To calculate this integral, we could expand that whole binomial, but it would be very time consuming and a bit of a pain. Instead, let's use u substitution:

Given this:

$\int5x^4(x^5-9)^6dx$

We can say that

Then, plug it back into our original expression

$\int u^6du$

Evaluate this integral to get

$\frac{u^7}{7}+c$

Then, replace u with what we substituted it for to get our final answer. Note because this is an indefinite integral, we need a plus c in it.

$\frac{(x^5-9)^7}{7}+c$

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Question

Integrate the following.

$\int \ \cos (2x+6) \ dx$

Answer

We can integrate using substitution:

and $du=2\ dx$ so $dx=\frac{1}{2} \ du$

$\int \ \cos (2x+6) \ dx= \frac{1}{2}\int \cos u \ du$

Now we can just focus on integrating cosine:

$\frac{1}{2}\int \cos u \ du = \frac{1}{2} \sin u +c$

Once the integration is complete, we can reinsert our substitution:

$\frac{1}{2} \sin (2x+6) +c$

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Question

Integrate the following.

$\int \sin (7x+3) \ dx$

Answer

We can integrate the function by using substitution where so $du=7 \ dx$ .

$\int \sin (7x+3) \ dx= \frac{1}{7}\int \sin u \ du$

Just focus on integrating sine now:

$\frac{1}{7}\int \sin u \ du = \frac{1}{7} (- \cos u +c)$

The last step is to reinsert the substitution:

$-\frac{1}{7} \cos \ (7x+3) +c$

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Question

Evaluate the following integral.

$\int \cos ^3 x\ dx$

Answer

Recall: The identity $\cos^2+\sin^2=1$

The integral can be rewritten as

$\int \cos ^3 x\ dx=\int \cos^2x \cdot \cos x \ dx$

Because of the trig identity above, we can rewrite it in a different way:

$\int \cos^2x \cdot \cos x \ dx = \int (1-\sin^2x ) \cos x \ dx$

Now we can integrate using substitution where $u=\sin x$ and $du= \cos x \ dx$

$\int (1-u^2 ) \ du=u-\frac{1}{3}u^3+c$

Finally, we reinsert our substitution:

$\sin x-\frac{1}{3}\sin ^3x+c$

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Question

Evaluate the following integral.

$\int \cos^4x\sin^3x\ dx$

Answer

Recall: The trig identity $\cos^2x+\sin^2x=1$

We can rewrite the integral using the above identity as

$\int \cos^4x\sin^3x\ dx=\int \cos^4x\sin^2x\ \sin x\ dx=\int \cos^4x(1-\cos ^2x) \sin x \ dx$

We can now solve the integral using substitution $u=\cos x$ and $du = -\sin x \ dx$

$- \int u^4(1-u ^2) du$ $=- \int u^4-u ^6 du$

$-\left(\frac{1}{5}u^5-\frac{1}{7}u ^7\right)$ $=-\frac{1}{5}u^5+\frac{1}{7}u ^7+c$

The last step is to reinsert our substitution:

$-\frac{1}{5}\cos ^5x+\frac{1}{7}\cos ^7x+c$

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Question

Fnd the derivative of tan(x) with respect to x or

$\frac{\mathrm{d} }{\mathrm{d} x} tan(x)dx$

Answer

The is one of the trigonometric integrals that must be memorized.

$\frac{\mathrm{d} }{\mathrm{d} x} tan(x)dx=sec^2(x)$

Other common trig derivatives that should be memorized are:

$\frac{\mathrm{d} }{\mathrm{d} x} sin(x)dx=cos(x)$

$\frac{\mathrm{d} }{\mathrm{d} x} cos(x)dx=-sin(x)$

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Question

Evaluate:

$\int_{0}^{\pi /3}\left ( \frac{cosx}{2}\right )dx$

Answer

The 1/2 is a constant, and so is pulled out front.

The integral of cos(x) is sin(x), by definition.

Writing the limits for evaluation:

$\frac{1}{2}sin(x)|^{\pi/3}_{0} = \frac{1}{2}\left [ sin(\pi/3)-sin(0)\right ]$

Using the unit circle, $sin(\pi/3)$ $=\sqrt3/2$ , and .

5)Simplifying:

$\frac{1}{2}\left [ \frac{\sqrt3}{2}-0\right ]=\frac{\sqrt3}{4}$

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