Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫₀¹/√2 2x arcsin(x²) dx

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Identify the integral to solve: $\int_0^{\frac{1}{\sqrt{2}}} 2x \arcsin(x^2) \, dx$.

Recognize that the integrand is a product of two functions: \$2x$ and $\arcsin(x^2)$. This suggests using integration by parts, where one function is differentiated and the other is integrated.

Choose $u = \arcsin(x^2)$, so that $du = \frac{d}{dx} \arcsin(x^2) \, dx$. Use the chain rule to find $du$: $du = \frac{1}{\sqrt{1 - (x^2)^2}} \cdot 2x \, dx = \frac{2x}{\sqrt{1 - x^4}} \, dx$.

Let $dv = 2x \, dx$, then integrate to find $v$: $v = \int 2x \, dx = x^2$.

Apply the integration by parts formula: $\int u \, dv = uv - \int v \, du$. Substitute the expressions for $u$, $v$, and $du$ to rewrite the integral and simplify before evaluating the definite integral from $0$ to $\frac{1}{\sqrt{2}}$.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is based on the product rule for differentiation and is given by ∫u dv = uv - ∫v du. Choosing u and dv wisely simplifies the integral, especially when one function becomes simpler upon differentiation.

Inverse Trigonometric Functions

Inverse trigonometric functions, like arcsin(x), are the inverses of the standard trigonometric functions. Understanding their derivatives and integrals is essential, as arcsin(x) has a derivative of 1/√(1 - x²), which often appears in integration problems involving these functions.

Definite Integrals and Limits of Integration

Definite integrals compute the net area under a curve between two limits. Evaluating a definite integral requires applying the Fundamental Theorem of Calculus after finding the antiderivative, then substituting the upper and lower limits to find the exact value.

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