Recognize the integral: Recognize the integral to be solved. We need to find the integral of the arcsin function with respect to t, which is written as ∫arcsin(t)dt.Use integration by parts: Use integration by parts. Integration by parts is given by the formula ∫u dv = uv - ∫v du, where u is a function of t, dv is the differential of another function of t, v is the integral of dv, and du is the differential of u. For the integral ∫arcsin(t)dt, we can let: u = arcsin(t) and dv = dt.Differentiate u and integrate dv: Differentiate u and integrate dv. Differentiating u with respect to t gives us: du = (1/√(1-t^2)) dt. Integrating dv with respect to t gives us: v = t.Apply the integration by parts formula: Apply the integration by parts formula. Now we apply the integration by parts formula: ∫arcsin(t)dt = t * arcsin(t) - ∫t * (1/√(1-t^2)) dt.Evaluate the remaining integral: Evaluate the remaining integral. The remaining integral ∫t * (1/√(1-t^2)) dt can be solved by a substitution. Let: w = 1 - t^2, then dw = -2t dt. Rearranging for dt, we get: dt = dw / (-2t). Substituting into the integral, we get: -1/2 ∫(1/√w) dw.Integrate with respect to w: Integrate with respect to w. The integral of 1/√w with respect to w is 2√w. So we have: -1/2 * 2√w = -√w. Substituting back for w, we get: -√(1 - t^2).Combine the results: Combine the results. Combining the results from integration by parts, we have: ∫arcsin(t)dt = t * arcsin(t) - (-√(1 - t^2)) + C, where C is the constant of integration. Simplifying, we get: ∫arcsin(t)dt = t * arcsin(t) + √(1 - t^2) + C.

Algebra 2

Sin, cos, and tan of special angles

Full solution

\int \arcsin t \, dt

Recognize the integral: Recognize the integral to be solved. $\newline$ We need to find the integral of the arcsin function with respect to $t$ , which is written as $\int \arcsin(t) \, dt$ .
Use integration by parts: Use integration by parts. $\newline$ Integration by parts is given by the formula $\int u \, dv = uv - \int v \, du$ , where $u$ is a function of $t$ , $dv$ is the differential of another function of $t$ , $v$ is the integral of $dv$ , and $du$ is the differential of $u$ . $\newline$ For the integral $\int \arcsin(t)\,dt$ , we can let: $\newline$ $u$ $0$ and $u$ $1$ .
Differentiate $u$ and integrate $dv$ : Differentiate $u$ and integrate $dv$ . Differentiating $u$ with respect to $t$ gives us: $du = (1/\sqrt{1-t^2}) dt$ . Integrating $dv$ with respect to $t$ gives us: $v = t$ .
Apply the integration by parts formula: Apply the integration by parts formula. $\newline$ Now we apply the integration by parts formula: $\newline$ $\int \arcsin(t)\,dt = t \cdot \arcsin(t) - \int t \cdot \left(\frac{1}{\sqrt{1-t^2}}\right) dt$ .
Evaluate the remaining integral: Evaluate the remaining integral. $\newline$ The remaining integral $\int t \cdot \left(\frac{1}{\sqrt{1-t^2}}\right) dt$ can be solved by a substitution. Let: $\newline$ $w = 1 - t^2$ , then $dw = -2t dt$ . $\newline$ Rearranging for $dt$ , we get: $\newline$ $dt = \frac{dw}{-2t}$ . $\newline$ Substituting into the integral, we get: $\newline$ $-\frac{1}{2} \int\left(\frac{1}{\sqrt{w}}\right) dw$ .
Integrate with respect to $w$ : Integrate with respect to $w$ . The integral of $\frac{1}{\sqrt{w}}$ with respect to $w$ is $2\sqrt{w}$ . So we have: $-\frac{1}{2} \times 2\sqrt{w} = -\sqrt{w}$ . Substituting back for $w$ , we get: $-\sqrt{1 - t^2}$ .
Combine the results: Combine the results. $\newline$ Combining the results from integration by parts, we have: $\newline$ $\int \arcsin(t)\,dt = t \cdot \arcsin(t) - (-\sqrt{1 - t^2}) + C$ , where $C$ is the constant of integration. $\newline$ Simplifying, we get: $\newline$ $\int \arcsin(t)\,dt = t \cdot \arcsin(t) + \sqrt{1 - t^2} + C$ .