# How Do You Solve the Integral of 1/(y+cos(x))^2?

• I
• Needhelp62
In summary, The integral of 1/(y+cos(x)) between x=0 and x=pi/2 can be found using the substitution t=tan(x/2) to be (2/sqrt(y^2-1))arctan(sqrt((y-1)/(y+1))). The integral of 1/(y+cos(x))^2 between x=0 and x=pi/2 cannot be solved using this substitution, but it can be expressed as a function of the original integral Q as (2-(2/(y+1))Q+(2/(y+1)).
Needhelp62
First part of the question was to work out the integral $$1/(y+cos(x))$$ between x=0 and x=pi/2 by using the substitution $$t=tan(x/2).$$
Got this to be $$\frac{2}{\sqrt{y^2-1}}arctan(\sqrt{\frac{y-1}{y+1}})$$
The next question says HENCE find integral with the same limits of $$\frac{1}{(y+cos(x))^2}$$
Ive tried using by parts to maybe get something minus the first integral but i can't get it to work. This exam is known for being full of horrible questions so anything is possible here

Hi, but ##y## is a fixed number or is a function ##y(x)##?

Ssnow said:
Hi, but ##y## is a fixed number or is a function ##y(x)##?
y is just a fixed number the question says the answer to the integral is a function of y

ok, in the risolution you must treat ##y## as a number and try with the substitution ##t=\tan{\frac{x}{2}}##. You will obtain an integral that is the ratio of two polynomials in ##t## ...

remember that ##\cos{x}=\frac{1-t^{2}}{1+t^{2}}##...

Ssnow said:
remember that ##\cos{x}=\frac{1-t^{2}}{1+t^{2}}##...
Thank you I get how the original integral i used now.I have two integrals with the first being twice the original and the second being
$$\frac{2t^2}{(y+1)+t^2(y-1)}$$

Needhelp62 said:
Thank you I get how the original integral i used now.I have two integrals with the first being twice the original and the second being
$$\frac{2t^2}{(y+1)+t^2(y-1)}$$
think I've got it, if i call the original integral Q so
$$(2-\frac{2}{y+1})Q+\frac{2}{y+1}$$

## What is the general formula for integrating 1/(y+cos(x))^2 dx?

The general formula for integrating 1/(y+cos(x))^2 dx is:

∫1/(y+cos(x))^2 dx = tan(x) + C

where C is the constant of integration.

## What is the method for solving the integral 1/(y+cos(x))^2 dx?

The method for solving the integral 1/(y+cos(x))^2 dx is to use the substitution technique. Let u = y + cos(x), then du = -sin(x)dx.

Rewriting the integral in terms of u, we get:

∫1/u^2 * (-1/sin(x)) du

= -∫1/u^2 * csc(x) du

= -cot(x) + C

= -cot(x) + C

= tan(x) + C

where C is the constant of integration.

## What is the domain of the function 1/(y+cos(x))^2?

The domain of the function 1/(y+cos(x))^2 is all real numbers except for the values of x that make the denominator, y + cos(x), equal to 0. This means that the domain is all real numbers except for the values of x where cos(x) = -y.

## What is the range of the function 1/(y+cos(x))^2?

The range of the function 1/(y+cos(x))^2 is all real numbers greater than or equal to 0. This is because the function is always positive, since the denominator, y + cos(x), is squared.

## How does the value of y affect the graph of the function 1/(y+cos(x))^2?

Changing the value of y will shift the graph of the function 1/(y+cos(x))^2 vertically. If y is positive, the graph will shift upwards, and if y is negative, the graph will shift downwards. This is because the value of y is added to the cosine function in the denominator, causing the graph to shift up or down.

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