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Chapter 1: Problem 3

Two of these three integrals are elementary; evaluate them (A) \(\int \sin ^{2} x d x\) (B) \(\int \sin \sqrt{x} d x\)\text { (C) } \int \sin x^{2} d x

Short Answer

Expert verified
Question: Identify the two elementary integrals and provide their solutions. Answer: The two elementary integrals are (A) \(\int \sin^2{x} dx\) and (B) \(\int \sin{(\sqrt{x})} dx\). Their solutions are: (A) \(\int \sin^2{x} dx = \frac{1}{2}(x - \frac{1}{2}\sin{2x}) + C\). (B) \(\int \sin{(\sqrt{x})} dx = -2x\cos{(\sqrt{x})} + 2\sin{(\sqrt{x})} + C\).

Step by step solution

01

Use the double-angle identity for cosine

We can rewrite \(\sin^2{x}\) using the double-angle identity for cosine: \(\sin^2{x} = \frac{1}{2}(1 - \cos{2x})\). Now, the integral becomes: \(\int \sin^2{x} dx = \int \frac{1}{2}(1 - \cos{2x}) dx\).
02

Integrate the function

Integrate the function term by term. \(\int \frac{1}{2}(1 - \cos{2x}) dx = \frac{1}{2} \int (1 - \cos{2x}) dx = \frac{1}{2} \int 1 dx - \frac{1}{2} \int \cos{2x} dx\).
03

Find the antiderivatives

Now, we need to find the antiderivatives for each integral. The antiderivative of \(1\) with respect to \(x\) is just \(x\), and the antiderivative of \(\cos{2x}\) with respect to \(x\) is \(\frac{1}{2}\sin{2x}\). So the overall antiderivative is: \(\frac{1}{2}(x - \frac{1}{2}\sin{2x}) + C\), where C is the constant of integration. Therefore, \(\int \sin^2{x} dx = \frac{1}{2}(x - \frac{1}{2}\sin{2x}) + C\). Integral (B): \(\int \sin{(\sqrt{x})} dx\)
04

Use substitution

We can use substitution to solve this integral. Let \(u = \sqrt{x}\), then \(x = u^2\), and \(dx = 2u du\). Now, we substitute these values in the integral, which becomes: \(\int \sin{(\sqrt{x})} dx = \int \sin{u} \cdot 2u du\).
05

Integrate using integration by parts

To integrate this, we can use integration by parts. Let \(dv = \sin{u} du\), and \(v = -\cos{u}\). Let \(u = u\), and \(du = du\). Therefore, by integration by parts, we have: \(\int u\sin{u} du = -u\cos{u} - \int(-\cos{u}) du = -u\cos{u} + \int \cos{u} du\) Now, the antiderivative of \(\cos{u}\) is \(\sin{u}\). So, the integral becomes: \(\int u\sin{u} du = -u\cos{u} + \sin{u} + C_1\)
06

Substitute back the original variable

Now, we need to substitute back the original variable \(x\). We know that \(u = \sqrt{x}\). So, we have: \(-u\cos{u} + \sin{u} + C_1 = -\sqrt{x}\cos{(\sqrt{x})} + \sin{(\sqrt{x})} + C_1\) Multiplying by the factor we got when substituting for \(x\), we obtain the final answer: \(\int \sin{(\sqrt{x})} dx = -2x\cos{(\sqrt{x})} + 2\sin{(\sqrt{x})} + C\). So, to summarize the solutions: Integral (A): \(\int \sin^2{x} dx = \frac{1}{2}(x - \frac{1}{2}\sin{2x}) + C\). Integral (B): \(\int \sin{(\sqrt{x})} dx = -2x\cos{(\sqrt{x})} + 2\sin{(\sqrt{x})} + C\).

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Most popular questions from this chapter

Evaluate the following integrals: (i) \(\int \frac{d x}{\sin x(3+2 \cos x)}\) (ii) \(\int \frac{\mathrm{d} \mathrm{x}}{\sin 2 \mathrm{x}-2 \sin \mathrm{x}}\) (iii) \(\int \frac{\sin \frac{\theta}{2} \tan \frac{\theta}{2} \mathrm{~d} \theta}{\cos \theta}\) (iv) \(\int \frac{d x}{\ln x^{x}\left[(\ln x)^{2}-3 \ln x-10\right]}\)

Evaluate the following integrals : (i) \(\int \frac{(\sqrt{x}+1)\left(x^{2}-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}} d x\) (ii) \(\int \frac{\sqrt{1-\mathrm{x}^{2}}-1}{\mathrm{x}}\left(\frac{1-\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}+\mathrm{x}-1}+\frac{\sqrt{1+\mathrm{x}}}{\sqrt{1+\mathrm{x}}-\sqrt{1-\mathrm{x}}}\right) \mathrm{dx}\) (iii) \(\int \frac{x^{4}+5 x^{3}+15 x-9}{\frac{x^{6}+3 x^{4}}+\frac{9}{x^{4}}}{\left(x^{3}-4 x+3 x^{2}-12\right) / x^{5}} d x\) (iv) \(\int \frac{\sqrt[3]{x+\sqrt{2-x^{2}}} \sqrt[6]{1-x \sqrt{2-x^{2}}}}{\sqrt[3]{1-x^{2}}} d x\)

Evaluate the following integrals : (i) \(\int \mathrm{e}^{\mathrm{x}}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}^{2}}\right)^{2} \mathrm{dx}\) (ii) \(\int \mathrm{e}^{x} \frac{\left(x^{3}-x+2\right)}{\left(x^{2}+1\right)^{2}} d x\) (iii) \(\int \frac{\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1)}{(\mathrm{x}+1)^{3}} \mathrm{dx}\) (iv) \(\int \mathrm{e}^{x}\left(\frac{1-x}{1+x}\right)^{2} d x\)

Evaluate the following integrals: (i) \(\int \frac{\mathrm{dx}}{\mathrm{x} \sqrt{\left(9 \mathrm{x}^{2}+4 \mathrm{x}+1\right)}}\) (ii) \(\int \frac{d x}{(1+x) \sqrt{\left(1+x-x^{2}\right)}}\) (iii) \(\int \frac{\mathrm{dx}}{(1+\mathrm{x}) \sqrt{1+2 \mathrm{x}-\mathrm{x}^{2}}}\) (iv) \(\int \frac{2 x d x}{\left(1-x^{2}\right) \sqrt{\left(x^{4}-1\right)}}\)

Evaluate the following integrals: (i) \(\int \frac{5 \cos ^{3} x+3 \sin ^{3} x}{\sin ^{2} x \cos ^{2} x} d x\) (ii) \(\int\left(\cos ^{6} x+\sin ^{6} x\right) d x\) (iii) \(\int \sin ^{3} x \cos \frac{x}{2} d x\) (iv) \(\int \frac{d x}{\sqrt{3} \cos x+\sin x}\)
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