Sine - GRE Quantitative Reasoning
Card 0 of 12
Evaluate:

Evaluate:
Evaluating this integral requires use of the "Product to Sum Formulas of Trigonometry":
For:
![\sin(\alpha x)\cos(\beta x)=\frac{1}{2}[\sin((\alpha - \beta)x)+\sin((\alpha + \beta)x)]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/426371/gif.latex)
So for our given integral, we can rewrite like so:
![\int \sin(75x)\cos(52x)dx\rightarrow \frac{1}{2}\int[\sin(23x)+\sin(127x)]dx](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/426372/gif.latex)
This can be rewritten as two separate integrals and solved using a simple substitution.
![\frac{1}{2}\int[\sin(23x)+\sin(127x)]dx\rightarrow \frac{1}{2}\int \sin(23x)dx+\frac{1}{2}\int\sin(127x)dx](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/426373/gif.latex)
Solving each integral individually, we have:


Substituting this into the integral results in:

The other integral is solved the same way:


Substituting this into the integral results in:

Now combining these two statements together results in one of the answer choices:

Evaluating this integral requires use of the "Product to Sum Formulas of Trigonometry":
For:
So for our given integral, we can rewrite like so:
This can be rewritten as two separate integrals and solved using a simple substitution.
Solving each integral individually, we have:
Substituting this into the integral results in:
The other integral is solved the same way:
Substituting this into the integral results in:
Now combining these two statements together results in one of the answer choices:
Compare your answer with the correct one above
Evaluate:

Evaluate:
This integral can be easily evaluated by following the rules outlined for integrating powers of sine and cosine.
But first a substitution needs to be made:



Now that we've made this substitution, we will use the rules outlined for integrating powers of sine and cosine:
In General:

1. If "m" is odd, then we make the substitution
, and we use the identity
.
2. If "n" is odd, then we make the substitution
, and we use the identity
.
For our given problem statement we will use the first rule, and alter the integral like so:

![\frac{1}{2}\int\sin^4(u)[\cos^2(u)]^2\cos(u)du\rightarrow\frac{1}{2}\int\sin^4(u)[1-\sin^2(u)]^2\cos(u)du](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/430221/gif.latex)

![\frac{1}{2}\int v^4[1-v^2]^2dv\rightarrow\frac{1}{2}\int v^4[1-2v^2+v^4]dv\rightarrow\frac{1}{2}\int (v^4-2v^6+v^8)dv](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/430223/gif.latex)
![\frac{1}{2}[\frac{1}{5}v^5-\frac{2}{7}v^7+\frac{1}{9}v^9]+C\rightarrow\frac{1}{10}v^5-\frac{1}{7}v^7+\frac{1}{18}v^9+C](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/430224/gif.latex)
Now we need to substitute back into v:

Now we need to substitute back into u, and rearrange to make it look like one of the answer choices:

This integral can be easily evaluated by following the rules outlined for integrating powers of sine and cosine.
But first a substitution needs to be made:
Now that we've made this substitution, we will use the rules outlined for integrating powers of sine and cosine:
In General:
1. If "m" is odd, then we make the substitution , and we use the identity
.
2. If "n" is odd, then we make the substitution , and we use the identity
.
For our given problem statement we will use the first rule, and alter the integral like so:
Now we need to substitute back into v:
Now we need to substitute back into u, and rearrange to make it look like one of the answer choices:
Compare your answer with the correct one above
Find
:

Find :
Step 1: Draw a
triangle..
The short sides have a length of
and the hypotenuse has a length of
.
Step 2: Find Sin (Angle A):

Step 3: Rationalize the root at the bottom:


Step 1: Draw a triangle..
The short sides have a length of and the hypotenuse has a length of
.
Step 2: Find Sin (Angle A):
Step 3: Rationalize the root at the bottom:
Compare your answer with the correct one above
Evaluate:

Evaluate:
Evaluating this integral requires use of the "Product to Sum Formulas of Trigonometry":
For:
![\sin(\alpha x)\cos(\beta x)=\frac{1}{2}[\sin((\alpha - \beta)x)+\sin((\alpha + \beta)x)]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/426371/gif.latex)
So for our given integral, we can rewrite like so:
![\int \sin(75x)\cos(52x)dx\rightarrow \frac{1}{2}\int[\sin(23x)+\sin(127x)]dx](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/426372/gif.latex)
This can be rewritten as two separate integrals and solved using a simple substitution.
![\frac{1}{2}\int[\sin(23x)+\sin(127x)]dx\rightarrow \frac{1}{2}\int \sin(23x)dx+\frac{1}{2}\int\sin(127x)dx](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/426373/gif.latex)
Solving each integral individually, we have:


Substituting this into the integral results in:

The other integral is solved the same way:


Substituting this into the integral results in:

Now combining these two statements together results in one of the answer choices:

Evaluating this integral requires use of the "Product to Sum Formulas of Trigonometry":
For:
So for our given integral, we can rewrite like so:
This can be rewritten as two separate integrals and solved using a simple substitution.
Solving each integral individually, we have:
Substituting this into the integral results in:
The other integral is solved the same way:
Substituting this into the integral results in:
Now combining these two statements together results in one of the answer choices:
Compare your answer with the correct one above
Evaluate:

Evaluate:
This integral can be easily evaluated by following the rules outlined for integrating powers of sine and cosine.
But first a substitution needs to be made:



Now that we've made this substitution, we will use the rules outlined for integrating powers of sine and cosine:
In General:

1. If "m" is odd, then we make the substitution
, and we use the identity
.
2. If "n" is odd, then we make the substitution
, and we use the identity
.
For our given problem statement we will use the first rule, and alter the integral like so:

![\frac{1}{2}\int\sin^4(u)[\cos^2(u)]^2\cos(u)du\rightarrow\frac{1}{2}\int\sin^4(u)[1-\sin^2(u)]^2\cos(u)du](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/430221/gif.latex)

![\frac{1}{2}\int v^4[1-v^2]^2dv\rightarrow\frac{1}{2}\int v^4[1-2v^2+v^4]dv\rightarrow\frac{1}{2}\int (v^4-2v^6+v^8)dv](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/430223/gif.latex)
![\frac{1}{2}[\frac{1}{5}v^5-\frac{2}{7}v^7+\frac{1}{9}v^9]+C\rightarrow\frac{1}{10}v^5-\frac{1}{7}v^7+\frac{1}{18}v^9+C](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/430224/gif.latex)
Now we need to substitute back into v:

Now we need to substitute back into u, and rearrange to make it look like one of the answer choices:

This integral can be easily evaluated by following the rules outlined for integrating powers of sine and cosine.
But first a substitution needs to be made:
Now that we've made this substitution, we will use the rules outlined for integrating powers of sine and cosine:
In General:
1. If "m" is odd, then we make the substitution , and we use the identity
.
2. If "n" is odd, then we make the substitution , and we use the identity
.
For our given problem statement we will use the first rule, and alter the integral like so:
Now we need to substitute back into v:
Now we need to substitute back into u, and rearrange to make it look like one of the answer choices:
Compare your answer with the correct one above
Find
:

Find :
Step 1: Draw a
triangle..
The short sides have a length of
and the hypotenuse has a length of
.
Step 2: Find Sin (Angle A):

Step 3: Rationalize the root at the bottom:


Step 1: Draw a triangle..
The short sides have a length of and the hypotenuse has a length of
.
Step 2: Find Sin (Angle A):
Step 3: Rationalize the root at the bottom:
Compare your answer with the correct one above
Evaluate:

Evaluate:
Evaluating this integral requires use of the "Product to Sum Formulas of Trigonometry":
For:
![\sin(\alpha x)\cos(\beta x)=\frac{1}{2}[\sin((\alpha - \beta)x)+\sin((\alpha + \beta)x)]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/426371/gif.latex)
So for our given integral, we can rewrite like so:
![\int \sin(75x)\cos(52x)dx\rightarrow \frac{1}{2}\int[\sin(23x)+\sin(127x)]dx](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/426372/gif.latex)
This can be rewritten as two separate integrals and solved using a simple substitution.
![\frac{1}{2}\int[\sin(23x)+\sin(127x)]dx\rightarrow \frac{1}{2}\int \sin(23x)dx+\frac{1}{2}\int\sin(127x)dx](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/426373/gif.latex)
Solving each integral individually, we have:


Substituting this into the integral results in:

The other integral is solved the same way:


Substituting this into the integral results in:

Now combining these two statements together results in one of the answer choices:

Evaluating this integral requires use of the "Product to Sum Formulas of Trigonometry":
For:
So for our given integral, we can rewrite like so:
This can be rewritten as two separate integrals and solved using a simple substitution.
Solving each integral individually, we have:
Substituting this into the integral results in:
The other integral is solved the same way:
Substituting this into the integral results in:
Now combining these two statements together results in one of the answer choices:
Compare your answer with the correct one above
Evaluate:

Evaluate:
This integral can be easily evaluated by following the rules outlined for integrating powers of sine and cosine.
But first a substitution needs to be made:



Now that we've made this substitution, we will use the rules outlined for integrating powers of sine and cosine:
In General:

1. If "m" is odd, then we make the substitution
, and we use the identity
.
2. If "n" is odd, then we make the substitution
, and we use the identity
.
For our given problem statement we will use the first rule, and alter the integral like so:

![\frac{1}{2}\int\sin^4(u)[\cos^2(u)]^2\cos(u)du\rightarrow\frac{1}{2}\int\sin^4(u)[1-\sin^2(u)]^2\cos(u)du](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/430221/gif.latex)

![\frac{1}{2}\int v^4[1-v^2]^2dv\rightarrow\frac{1}{2}\int v^4[1-2v^2+v^4]dv\rightarrow\frac{1}{2}\int (v^4-2v^6+v^8)dv](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/430223/gif.latex)
![\frac{1}{2}[\frac{1}{5}v^5-\frac{2}{7}v^7+\frac{1}{9}v^9]+C\rightarrow\frac{1}{10}v^5-\frac{1}{7}v^7+\frac{1}{18}v^9+C](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/430224/gif.latex)
Now we need to substitute back into v:

Now we need to substitute back into u, and rearrange to make it look like one of the answer choices:

This integral can be easily evaluated by following the rules outlined for integrating powers of sine and cosine.
But first a substitution needs to be made:
Now that we've made this substitution, we will use the rules outlined for integrating powers of sine and cosine:
In General:
1. If "m" is odd, then we make the substitution , and we use the identity
.
2. If "n" is odd, then we make the substitution , and we use the identity
.
For our given problem statement we will use the first rule, and alter the integral like so:
Now we need to substitute back into v:
Now we need to substitute back into u, and rearrange to make it look like one of the answer choices:
Compare your answer with the correct one above
Find
:

Find :
Step 1: Draw a
triangle..
The short sides have a length of
and the hypotenuse has a length of
.
Step 2: Find Sin (Angle A):

Step 3: Rationalize the root at the bottom:


Step 1: Draw a triangle..
The short sides have a length of and the hypotenuse has a length of
.
Step 2: Find Sin (Angle A):
Step 3: Rationalize the root at the bottom:
Compare your answer with the correct one above
Evaluate:

Evaluate:
Evaluating this integral requires use of the "Product to Sum Formulas of Trigonometry":
For:
![\sin(\alpha x)\cos(\beta x)=\frac{1}{2}[\sin((\alpha - \beta)x)+\sin((\alpha + \beta)x)]](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/426371/gif.latex)
So for our given integral, we can rewrite like so:
![\int \sin(75x)\cos(52x)dx\rightarrow \frac{1}{2}\int[\sin(23x)+\sin(127x)]dx](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/426372/gif.latex)
This can be rewritten as two separate integrals and solved using a simple substitution.
![\frac{1}{2}\int[\sin(23x)+\sin(127x)]dx\rightarrow \frac{1}{2}\int \sin(23x)dx+\frac{1}{2}\int\sin(127x)dx](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/426373/gif.latex)
Solving each integral individually, we have:


Substituting this into the integral results in:

The other integral is solved the same way:


Substituting this into the integral results in:

Now combining these two statements together results in one of the answer choices:

Evaluating this integral requires use of the "Product to Sum Formulas of Trigonometry":
For:
So for our given integral, we can rewrite like so:
This can be rewritten as two separate integrals and solved using a simple substitution.
Solving each integral individually, we have:
Substituting this into the integral results in:
The other integral is solved the same way:
Substituting this into the integral results in:
Now combining these two statements together results in one of the answer choices:
Compare your answer with the correct one above
Evaluate:

Evaluate:
This integral can be easily evaluated by following the rules outlined for integrating powers of sine and cosine.
But first a substitution needs to be made:



Now that we've made this substitution, we will use the rules outlined for integrating powers of sine and cosine:
In General:

1. If "m" is odd, then we make the substitution
, and we use the identity
.
2. If "n" is odd, then we make the substitution
, and we use the identity
.
For our given problem statement we will use the first rule, and alter the integral like so:

![\frac{1}{2}\int\sin^4(u)[\cos^2(u)]^2\cos(u)du\rightarrow\frac{1}{2}\int\sin^4(u)[1-\sin^2(u)]^2\cos(u)du](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/430221/gif.latex)

![\frac{1}{2}\int v^4[1-v^2]^2dv\rightarrow\frac{1}{2}\int v^4[1-2v^2+v^4]dv\rightarrow\frac{1}{2}\int (v^4-2v^6+v^8)dv](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/430223/gif.latex)
![\frac{1}{2}[\frac{1}{5}v^5-\frac{2}{7}v^7+\frac{1}{9}v^9]+C\rightarrow\frac{1}{10}v^5-\frac{1}{7}v^7+\frac{1}{18}v^9+C](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/430224/gif.latex)
Now we need to substitute back into v:

Now we need to substitute back into u, and rearrange to make it look like one of the answer choices:

This integral can be easily evaluated by following the rules outlined for integrating powers of sine and cosine.
But first a substitution needs to be made:
Now that we've made this substitution, we will use the rules outlined for integrating powers of sine and cosine:
In General:
1. If "m" is odd, then we make the substitution , and we use the identity
.
2. If "n" is odd, then we make the substitution , and we use the identity
.
For our given problem statement we will use the first rule, and alter the integral like so:
Now we need to substitute back into v:
Now we need to substitute back into u, and rearrange to make it look like one of the answer choices:
Compare your answer with the correct one above
Find
:

Find :
Step 1: Draw a
triangle..
The short sides have a length of
and the hypotenuse has a length of
.
Step 2: Find Sin (Angle A):

Step 3: Rationalize the root at the bottom:


Step 1: Draw a triangle..
The short sides have a length of and the hypotenuse has a length of
.
Step 2: Find Sin (Angle A):
Step 3: Rationalize the root at the bottom:
Compare your answer with the correct one above