Chapter 6: Problem 15
In Exercises \(11-20,\) solve the initial value problem explicitly. $$\frac{d y}{d x}=-\frac{1}{x^{2}}-\frac{3}{x^{4}}+12\( and \)y=3\( when \)x=1$$
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\(\int \sec x d x \quad\) (Hint: Multiply the integrand by \(\frac{\sec x+\tan x}{\sec x+\tan x}\) and then use a substitution to integrate the result.)
Trigonometric Substitution Suppose \(u=\sin ^{-1} x .\) Then \(\cos u>0\) . (a) Use the substitution \(x=\sin u, d x=\cos u d u\) to show that $$\int \frac{d x}{\sqrt{1-x^{2}}}=\int 1 d u$$ (b) Evaluate \(\int 1 d u\) to show that \(\int \frac{d x}{\sqrt{1-x^{2}}}=\sin ^{-1} x+C\)
Second-Order Differential Equations Find the specific solution to each of the following second-order initial value problems by first finding \(d y / d x\) and then finding \(y\) . (a) \(\frac{d^{2} y}{d x^{2}}=24 x^{2}-10\) when \(x=1, \frac{d y}{d x}=3\) and \(y=5\) (b) \(\frac{d^{2} y}{d x^{2}}=\cos x-\sin x when \)x=0, \frac{d y}{d x}=2\( and \)y=0\( (c) \)\frac{d^{2} y}{d x^{2}}=e^{x}-x\( when \)x=0, \frac{d y}{d x}=0\( and \)y=1$
Finding Area Find the area of the region enclosed by the \(x\) -axis and the curve \(y=x \sin x\) for (a) \(0 \leq x \leq \pi\) (b) \(\pi \leq x \leq 2 \pi\) (c) \(0 \leq x \leq 2 \pi\)
In Exercises \(53-66,\) make a \(u\) -substitution and integrate from \(u(a)\) to \(u(b) .\) $$\int_{0}^{3} \sqrt{y+1} d y$$
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