Step-by-step explanation:
To integrate [tex]\( \frac{\cos x}{\sin^2 x + 1} \)[/tex], we can use a substitution method. Let's proceed step by step:
1. Substitution: Use [tex]\( u = \sin x \)[/tex], hence [tex]\( du = \cos x \, dx \)[/tex].
2. Rewrite the Integral: Substitute [tex]\( u = \sin x \)[/tex] and [tex]\( du = \cos x \, dx \)[/tex]:
[tex]\[ \int \frac{\cos x}{\sin^2 x + 1} \, dx = \int \frac{1}{u^2 + 1} \, du \][/tex]
3. Integrate: Now integrate [tex]\( \int \frac{1}{u^2 + 1} \, du \)[/tex]. The antiderivative of [tex]\( \frac{1}{u^2 + 1} \)[/tex] is [tex]\( \arctan(u) \)[/tex]:
[tex]\[ \int \frac{1}{u^2 + 1} \, du = \arctan(u) + C \][/tex]
4. Back Substitute: Substitute back [tex]\( u = \sin x \)[/tex]:
[tex]\[ \int \frac{\cos x}{\sin^2 x + 1} \, dx = \arctan(\sin x) + C \][/tex]
Therefore, the integral [tex]\( \int \frac{\cos x}{\sin^2 x + 1} \, dx \)[/tex] evaluates to [tex]\( \arctan(\sin x) + C \)[/tex].
Answer:
To integrate \( \frac{\cos x}{\sin^2 x + 1} \), let's proceed with the solution step by step:
1. **Rewrite the denominator** using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \):
\[ \sin^2 x + 1 = \sin^2 x + \cos^2 x + 1 - \cos^2 x = 1 - \cos^2 x + 1 = 2 - \cos^2 x \]
2. Substitute this into the original integral:
\[ \int \frac{\cos x}{\sin^2 x + 1} \, dx = \int \frac{\cos x}{2 - \cos^2 x} \, dx \]
3. Use the substitution \( u = \sin x \), \( du = \cos x \, dx \):
\[ \int \frac{du}{2 - (1 - u^2)} = \int \frac{du}{1 + u^2} \]
4. Integrate \( \frac{1}{1 + u^2} \) with respect to \( u \):
\[ \int \frac{du}{1 + u^2} = \arctan(u) + C \]
5. Substitute back \( u = \sin x \):
\[ \int \frac{\cos x}{\sin^2 x + 1} \, dx = \arctan(\sin x) + C \]
Therefore, the integral of \( \frac{\cos x}{\sin^2 x + 1} \) is \( \arctan(\sin x) + C \), where \( C \) is the constant of integration.