Evaluating Indefinite Integrals

Evaluate the integrals in Exercises 37–46.

∫ 2(cos x)⁻¹/² sin x dx

Guida verificata passo dopo passo

Recognize that the integral is of the form \(\int 2 (\cos x)^{-\frac{1}{2}} \sin x \, dx\), which suggests a substitution involving \(\cos x\) because its derivative is related to \(\sin x\).

Let \(u = \cos x\). Then, compute the differential \(du = -\sin x \, dx\), which implies \(-du = \sin x \, dx\).

Rewrite the integral in terms of \(u\): substitute \(\cos x\) with \(u\) and \(\sin x \, dx\) with \(-du\), so the integral becomes \(\int 2 u^{-\frac{1}{2}} (-du)\).

Simplify the integral to \(-2 \int u^{-\frac{1}{2}} \, du\), which is a basic power integral.

Use the power rule for integration: \(\int u^{n} \, du = \frac{u^{n+1}}{n+1} + C\) for \(n \neq -1\), and apply it to \(u^{-\frac{1}{2}}\) to find the antiderivative.

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