Trigonometric Integrals - GRE Quantitative Reasoning

Card 0 of 24

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:

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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:

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

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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:

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

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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:

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

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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:

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

Compare your answer with the correct one above

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