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GRE Subject Test: Math : Integrals

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Example Questions

Example Question #3 : Trigonometric Integrals

Evaluate the following integral.

$\int \cos ^3 x\ dx$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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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Example Question #4 : Trigonometric Integrals

Evaluate the following integral.

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

Possible Answers:

$-\frac{1}{5}\cos ^5x\sin ^3x+\frac{1}{4}\sin ^4x\cos ^4x+c$

$\frac{1}{5}\cos ^5x\sin ^3x-\frac{1}{4}\sin ^4x\cos ^4x+c$

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

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

Correct answer:

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

Explanation:

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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Example Question #11 : Integrals

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

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

Possible Answers:

cos^2(x)

Derivative cannot be found

sec^2(x)

cos(x)

sec(x)

Correct answer:

sec^2(x)

Explanation:

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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Example Question #1 : Trigonometric Integrals

Evaluate:

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

Possible Answers:

$\frac{\sqrt3-4}{4}$

$\frac{\pi}{3}$

$\frac{\sqrt3}{2}$

$\frac{\sqrt3}{4}$

$\frac{1}{4}$

Correct answer:

$\frac{\sqrt3}{4}$

Explanation:

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

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

3) Writing the limits for evaluation:

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

4) Using the unit circle, $sin(\pi/3)$ $=\sqrt3/2$ , and sin(0)=0 .

5)Simplifying:

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

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GRE Subject Test: Math : Integrals

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #3 : Trigonometric Integrals

Example Question #4 : Trigonometric Integrals

Example Question #11 : Integrals

Example Question #1 : Trigonometric Integrals

All GRE Subject Test: Math Resources

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