# Finding an integral using trig. substitution

In summary, the integral from 0 to pi/2 of cos(t)/sqrt(1+sin^2(t)) dt can be solved using trigonometric substitution. By substituting sin(t)=x, the integral becomes 1/sqrt(1+x^2)dx. Using a trigonometric substitution, the integral can be simplified to sec(θ)dθ. However, when evaluating the integral at pi/2, it results in infinity due to the secant and tangent values being infinite at that point. Further adjustments and translations of the upper and lower limits are needed to obtain the correct solution.
The integral from 0 to pi/2 of:

cos(t)/sqrt(1+sin^2(t)) dt

I'm supposed to use trig. substitution to find the solution. I started by using the formula a^2+x^2 to get x=atanx. In this case, sin(t)=(1)tan(θ), and so cos(t)dt=sec^2(θ)dθ and so I substituted this into the equation and got:

sec^2(θ)/sqrt(1+tan^2(θ)) -> sec^2(θ)/sqrt(sec^2(θ)) -> sec(θ)

Now, I have the integral of sec(θ)dθ, which equals ln abs(secθ+tanθ). When I take this integral at 0, it turns out to be 0, and so I'm left with the answer being solely the integral at pi/2. The problem is that the secant at pi/2 is infinity and so is the tangent at pi/2, and so the answer ends up being infinity, and this is apparently wrong.

Any help would be greatly appreciated.

It would be better if you substitute sin(t)=x.

A trig substitution can be used after you use the substitution that Pranav-Arora recommends.

Pranav-Arora said:
It would be better if you substitute sin(t)=x.

I substituted sin(t)=x and got:

1/sqrt(1+x^2)dx

but when I use the trig. substitution with this integral, I come up with the same thing I had before. I don't think I understand how to do it substituting sin(t)=x and then using a trig. substitution, like Mark said.

Never mind. I eventually figured out how to translate the upper and lower limits of the integrand over to the final versions. Thanks.

## 1. How do I know when to use trigonometric substitution?

Trigonometric substitution is used when the integrand (the expression being integrated) involves a radical or a quadratic equation that cannot be easily simplified. The substitution allows us to convert the integral into a form that can be solved using trigonometric identities.

## 2. What are the three types of trigonometric substitution?

The three types of trigonometric substitution are:

• Type 1: Substituting x with a sin θ
• Type 2: Substituting x with a sec θ
• Type 3: Substituting x with a tan θ

## 3. How do I choose which type of trigonometric substitution to use?

The type of trigonometric substitution to use depends on the form of the integrand. For integrands involving sqrt(a2 - x2), use Type 1. For integrands involving sqrt(a2 + x2), use Type 2. For integrands involving sqrt(x2 - a2), use Type 3.

## 4. What are the common trigonometric identities used in trigonometric substitution?

The common trigonometric identities used in trigonometric substitution are:

• Sine: sin2θ + cos2θ = 1
• Cosine: 1 + tan2θ = sec2θ
• Tangent: 1 + cot2θ = csc2θ

## 5. What are some tips for solving integrals using trigonometric substitution?

Some tips for solving integrals using trigonometric substitution are:

• Always draw a right triangle and label the sides using the given information.
• Choose the appropriate type of trigonometric substitution based on the form of the integrand.
• Use the common trigonometric identities to simplify the integrand.
• Remember to substitute back x in terms of θ and include the correct limits of integration.

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