int_(0)^(pi)sin(3x+(pi)/(2))dx

Apply Trigonometric Identity: We need to evaluate the integral of sin(3x + (pi/2)) from 0 to pi. To simplify the integral, we can use the trigonometric identity sin(a + b) = sin(a)cos(b) + cos(a)sin(b). In this case, a = 3x and b = (pi/2), so sin(b) = 1 and cos(b) = 0. This simplifies the integral to just the integral of sin(3x).Use Substitution Method: Now we can write the integral as: int_(0)^(pi)sin(3x)dx To integrate sin(3x), we use the substitution u = 3x, which implies du = 3dx or dx = du/3.Change Limits of Integration: Substituting u and dx into the integral, we get: int_(0)^(3pi)sin(u)(du/3) The limits of integration have also changed because when x = 0, u = 0, and when x = pi, u = 3pi.Evaluate Integral of sin(u): The integral now becomes: (1/3)int_(0)^(3pi)sin(u)du The integral of sin(u) with respect to u is -cos(u), so we have: (1/3)(-cos(u)) evaluated from 0 to 3pi.Final Simplification: Evaluating the antiderivative at the bounds, we get: (1/3)(-cos(3pi)) - (1/3)(-cos(0)) We know that cos(3pi) = -1 and cos(0) = 1, so the expression simplifies to: (1/3)(1) - (1/3)(-1) = (1/3) + (1/3) = 2/3.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

$\int_{0}^{\pi}\sin(3x+\frac{\pi}{2})dx$

View step-by-step help

Home

Math Problems

Calculus

Evaluate definite integrals using the chain rule

Full solution

\int_{0}^{\pi}\sin(3x+\frac{\pi}{2})dx

Apply Trigonometric Identity: We need to evaluate the integral of $\sin(3x + (\pi/2))$ from $0$ to $\pi$ . To simplify the integral, we can use the trigonometric identity $\sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b)$ . In this case, $a = 3x$ and $b = (\pi/2)$ , so $\sin(b) = 1$ and $\cos(b) = 0$ . This simplifies the integral to just the integral of $\sin(3x)$ .
Use Substitution Method: Now we can write the integral as: $\newline$ $\int_{0}^{\pi}\sin(3x)\,dx$ $\newline$ To integrate $\sin(3x)$ , we use the substitution $u = 3x$ , which implies $du = 3dx$ or $dx = \frac{du}{3}$ .
Change Limits of Integration: Substituting $u$ and $dx$ into the integral, we get: $\newline$ $\int_{0}^{3\pi}\sin(u)\left(\frac{du}{3}\right)$ $\newline$ The limits of integration have also changed because when $x = 0$ , $u = 0$ , and when $x = \pi$ , $u = 3\pi$ .
Evaluate Integral of $\sin(u)$ : The integral now becomes: $(\frac{1}{3})\int_{0}^{3\pi}\sin(u)\,du$ The integral of $\sin(u)$ with respect to $u$ is $-\cos(u)$ , so we have: $(\frac{1}{3})(-\cos(u))$ evaluated from $0$ to $3\pi$ .
Final Simplification: Evaluating the antiderivative at the bounds, we get: $\newline$ $(\frac{1}{3})(-\cos(3\pi)) - (\frac{1}{3})(-\cos(0))$ $\newline$ We know that $\cos(3\pi) = -1$ and $\cos(0) = 1$ , so the expression simplifies to: $\newline$ $(\frac{1}{3})(1) - (\frac{1}{3})(-1) = (\frac{1}{3}) + (\frac{1}{3}) = \frac{2}{3}$ .