Find:
380
(b) All the antiderivatives of \(f(x) =\sqrt{x} =x^{1/2} \) are \(F(x) =\dfrac{x^{3/2}}{\dfrac{3}{2}}+C=\dfrac{2x^{3/2}}{3}+C\), so \[ \int \sqrt{x} dx=\dfrac{2x^{3/2}}{3}+C \]
Trigonometric identities are discussed in Appendix A.4, pp. A-32 to A-35.
(c) No antiderivative in Table 1 corresponds to \(f(x)=\dfrac{\sin x}{\cos ^{2}x}\), so we begin by using trigonometric identities to rewrite \(\dfrac{\sin x}{\cos ^{2}x}\) in a form whose antiderivative is recognizable. \[ \dfrac{\sin x}{\cos ^{2}x}=\dfrac{\sin x}{\cos x\cdot \cos x}=\dfrac{1}{\cos x}\cdot \dfrac{\sin x}{\cos x}=\sec x\tan x \]
Then \[ \int \dfrac{\sin x}{\cos ^{2}x}\, dx=\int \sec x\tan x dx=\sec x+C \]