IDNLearn.com provides a user-friendly platform for finding answers to your questions. Ask your questions and receive reliable, detailed answers from our dedicated community of experts.
Sagot :
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.
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to assisting you again.
