Introduction to integral of sin2x/1+cos2x

In calculus, the integral is a fundamental concept that assigns numbers to functions to define displacement, area, volume, and all those functions that contain a combination of tiny elements. It is categorized into two parts, definite integral and indefinite integral. The process of integration calculates the integrals. This process is defined as finding an antiderivative of a function.

Integrals can handle almost all functions, such as trigonometric, algebraic, exponential, logarithmic, etc. This article will teach you what is integral to a trigonometric function sine. You will also understand how to compute sin2x by 1+cos2x integral by using different integration techniques.

What is the integral of sin2x/1+cos2x?

The integral of sin2x/1+cos2x is an antiderivative of sine function which is equal to –ln|1+cos^2x|+c. It is also known as the reverse derivative of sine function which is a trigonometric identity.

The sine function is the ratio of opposite side to the hypotenuse of a triangle which is written as:

Sin = opposite side / hypotenuse

Integral of sin2x/1+cos²x formula

The formula of integral of sin contains integral sign, coefficient of integration and the function as sine. It is denoted by ∫(sin2x/sin x)dx. In mathematical form, the integral of sin2x/1+cos2x is:

$∫\frac{\sin2x}{1+\cos^2x}dx=-\ln|1+\cos^2x| + c$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of integral. Integral of Sin x / Cos^2x can also be calculated by using the above integration formula.

How to calculate the integral of sin(2x)/1+cos^2x?

The integral of sin2x/sin x is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate integral of sine by using:

1. Substitution method
2. Definite integral

Integral of sin2x/1+cos2x by using substitution method

The substitution method involves many trigonometric formulas. We can use these formulas to verify the integrals of different trigonometric functions such as sine, cosine, tangent, etc. Let’s understand how to prove the integral of sin2x/1+cos^2x by using the substitution method.

Proof of Integral of sin2x/1+cos2x by using substitution method

To proof the integral of sin2x/1+cos2x by using substitution method, suppose that:

$y = \frac{\sin2x}{1+\cos^2x}$

Applying integral,

$I = ∫\frac{\sin2x}{1+\cos^2x}.dx$

Since we know that,

$\Sin2x = 2\sinx.\cosx$

Therefore assume that

$u = \cos x\quad\tex{and}\quad du =-\sinx dx$

Since we are using a parameter u to rewrite the given integral, therefore, this method is known as u-substitution. Using this substitution in the integral,

$I = ∫\frac{-2u}{1+u^2}.du$

Since

$\frac{d}{dx}(1+u^2) = 2u$

Now integrating with respect to u we get,

$I = -\ln|1+u^2| + c$

Now substituting the value of u here,

$I = -\ln|1+\cos^2x|+c$

Hence we have verified the integral of sin2x/1+cos^2x by using substitution method.

Integral of sin2x/1+cos^2x by using definite integral

The definite integral is a type of integral that calculates the area of a curve by using infinitesimal area elements between two points. The definite integral can be written as:

$∫^b_a f(x) dx = F(b) – F(a)

Let’s understand the verification of the integral of sin x by using the definite integral.

Proof of integral of sin2x/1 + cos2x by using definite integral

To compute the integral of sin x by using a definite integral calculator, we can use the interval from 0 to π or 0 to π/2. Let’s compute the integral of sin x from 0 to π. For this we can write the integral as:

$∫^π_0 \frac{\sin2x}{1 + \cos2x}.dx=-|\ln|1+\cos^2x||^π_0$

Now, substituting the limit in the given function.

$∫^π_0 \frac{\sin2x}{1 + \cos2x}.dx=-\ln|1+\cos^2(π)| +\ln|1+\cos^2(0)|$

Since cos 0 is equal to 1 and cos π is equal to -1, therefore,

$∫^π_0 \frac{\sin2x}{1 + \cos2x}.dx = 0 +\ln(2) =0.7$

Which is the calculation of the definite integral of sin2x/1+cos2x. Now to calculate the integral of sin x between the interval 0 to π/2, we just have to replace π by π/2. Therefore,

$∫^{\frac{π}{2}}_0 \frac{\sin2x}{1+\cos^2x}.dx =-|\ln|1+\cos^2x||^{\frac{π}{2}}_0$

Now, substituting the limit in the given function.

$∫^{\frac{π}{2}}_0 \frac{\sin2x}{1+\cos^2x}.dx=-\ln|1+\cos^2\frac{π}{2}|+\ln|1+\cos^2(0)|$

Since cos 0 is equal to 1 and cos π is equal to -1, therefore,

$∫^{\frac{π}{2}}_0 \frac{\sin2x}{1+\cos^2x}.dx = 0 + \ln(2) = 0.7$

Therefore, the definite integral of sin2x/1+cos2x is equal to 0.7.