Answer to Question #207317 in Real Analysis for adreanna

Question

Calculate the following integrals by using the integration

methods.

1

a) ∫ (4𝑥^34 + 8𝑥^3 + 15𝑥)𝑑𝑥 / (√𝑥^2 + 4x)

0

b) ∫ 𝑑𝑥 / sin 𝑥 ∙ cos^2 𝑥

Answer 1

(a)

$u=x+2, du=dx$

$4𝑥^4 + 8𝑥^3 + 15𝑥=4x^3(x+2)+15x$

$=4u(u-2)^3+15(u-2)$

$=4u^4-24u^3+48u^2-32u+15u-30$

$=4u^2(u^2-4)+16u^2-24u(u^2-4)-96u$

$+48u^2-32u+15u-30$

$=4u^2(u^2-4)-24u(u^2-4)+64(u^2-4)$

$-113u+226$

$\sqrt{x^2+4x}=\sqrt{x^2+4x+4-4}= \sqrt{u^2-4}$

Table of Integrals

$\int u^2\sqrt{u^2-a^2}du=\dfrac{u}{8}(2u^2-a^2)\sqrt{u^2-a^2}$

$-\dfrac{a^4}{8}\ln|u+\sqrt{u^2-a^2}|+C$

$\int u\sqrt{u^2-a^2}du=\dfrac{1}{3}(u^2-a^2)\sqrt{u^2-a^2}+C$

$-\dfrac{a^4}{8}\ln|u+\sqrt{u^2-a^2}|+C$

$\int \sqrt{u^2-a^2}du=\dfrac{u}{2}\sqrt{u^2-a^2}$

$-\dfrac{a^2}{2}\ln|u+\sqrt{u^2-a^2}|+C$

$\int\dfrac{u}{\sqrt{u^2-a^2}}du=\sqrt{u^2-a^2}+C$

$\int\dfrac{1}{\sqrt{u^2-a^2}}du=\ln|u+\sqrt{u^2-a^2}|+C$

$\int\dfrac{4u^4-24u^3+48u^2-17u-30}{\sqrt{u^2-4}}du$

$=\int4u^2\sqrt{u^2-4}du-\int24u\sqrt{u^2-4}du$

$+\int64\sqrt{u^2-4}du-\int\dfrac{113u}{\sqrt{u^2-4}}du+\int\dfrac{226}{\sqrt{u^2-4}}du$

$=u(u^2-2)\sqrt{u^2-4}-8\ln|u+\sqrt{u^2-4}|-$

$-8(u^2-4)\sqrt{u^2-4}$

$+32u\sqrt{u^2-4}-128\ln|u+\sqrt{u^2-4}|$

$-113\sqrt{u^2-4}+226\ln|u+\sqrt{u^2-4}|+C$

$=\sqrt{u^2-4}(u^3-2u-8u^2+32+32u-113)$

$+90\ln|u+\sqrt{u^2-4}|+C$

$=\sqrt{x^2+4x}((x+2)^3-8(x+2)^2+30(x+2)-81)$

$+90\ln|x+2+\sqrt{x^2+4x}|+C$

$\displaystyle\int_{0}^{1}\dfrac{4𝑥^4 + 8𝑥^3 + 15𝑥}{ \sqrt{x^2+4x}}dx$

$=\sqrt{5}(27-72+90-81)+90\ln(3+\sqrt{5})$

$-90\ln(2)=90\ln(\dfrac{3+\sqrt{5}}{2})-36$

(b)

$\int\dfrac{dx}{\sin x\cos^2 x}=\int\dfrac{(1+\tan^2 x)dx}{\sin x}$

$\int\dfrac{dx}{\sin x}=\int\dfrac{\sin x dx}{\sin^2 x}=\int\dfrac{\sin xdx}{1-\cos^2 x}$

$=\dfrac{1}{2}\int\dfrac{\sin xdx}{1-\cos x}-\dfrac{1}{2}\int\dfrac{\sin xdx}{1+\cos x}$

$=\dfrac{1}{2}\ln(1-\cos x)-\dfrac{1}{2}\ln(1+\cos x)+C_1$

$\int\dfrac{\tan^2 xdx}{\sin x}=\int\dfrac{\sin x dx}{\cos^2 x}=\dfrac{1}{\cos x}+C_2$

Therefore

$\int\dfrac{dx}{\sin x\cos^2 x}=\dfrac{1}{2}\ln(1-\cos x)-\dfrac{1}{2}\ln(1+\cos x)$

$+\dfrac{1}{\cos x}+C$