Definite Integrals — Questions 154 to 204 (Step-by-Step Solutions)

Part 1: Q.154–174

154. 154. \(I=\displaystyle\int_{0}^{\pi} x\,f(\sin x)\,dx.\)
     Let \(I=\int_0^\pi x f(\sin x)\,dx\).

     Substitute \(x\mapsto\pi-x\):

     \[ I=\int_0^\pi (\pi-x) f(\sin(\pi-x))\,dx =\int_0^\pi (\pi-x) f(\sin x)\,dx. \]

     Add the two expressions:

     \[ 2I=\pi\int_0^\pi f(\sin x)\,dx \quad\Rightarrow\quad I=\frac{\pi}{2}\int_0^\pi f(\sin x)\,dx. \]
155. 155. Show \(\displaystyle\int_{0}^{\pi} x f(\sin x)\,dx=\pi\int_{0}^{\pi/2} f(\sin x)\,dx.\)
     From Q.154, \(I=\dfrac{\pi}{2}\int_0^\pi f(\sin x)\,dx\). But \(\int_0^\pi f(\sin x)\,dx=2\int_0^{\pi/2} f(\sin x)\,dx\) since \(\sin(\pi-x)=\sin x\). Therefore \(I=\pi\int_0^{\pi/2} f(\sin x)\,dx\).
156. 156. Show \(f(\cos x)\) is even and relate integrals.
     \(\cos(-x)=\cos x\Rightarrow f(\cos x)\) even. So \(\int_{-\pi/2}^{\pi/2} f(\cos x)\,dx=2\int_0^{\pi/2} f(\cos x)\,dx\). Use \(t=\tfrac{\pi}{2}-x\) to convert \(\int_0^{\pi/2} f(\cos x)\,dx=\int_0^{\pi/2} f(\sin t)\,dt\).
157. 157. \(f(t)=\log\big(t+\sqrt{1+t^2}\big)\). Show \(f\) is odd; deduce \(\phi(x)=\int_a^x f(t)\,dt\) is even.
     \[ f(-t)=\log(-t+\sqrt{1+t^2}). \] Note \((t+\sqrt{1+t^2})(-t+\sqrt{1+t^2})=1\) so \(f(-t)=-f(t)\). Hence \(f\) odd and \(\phi(x)\) even.
158. 158. \(f(x)=\displaystyle\int_{0}^{x}\log\frac{1-t}{1+t}\,dt\). Show \(f\) even.
     Replace \(t=-u\): \[ f(-x)=\int_0^{-x}\log\frac{1-t}{1+t}\,dt=\int_0^x\log\frac{1-u}{1+u}\,du=f(x). \]
159. 159. If \(F(x)=\int_a^x f(t)\,dt\) and \(f\) odd, show \(F\) even.
     \(F(-x)=\int_a^{-x} f(t)\,dt\). Put \(t=-u\) and use \(f(-u)=-f(u)\) to obtain \(F(-x)=F(x)\).
160. 160. \(f(x)=\sin^3x\cos^2x\). Show odd and evaluate integral on \([-\pi,\pi]\).
     \(\sin^3(-x)=-\sin^3x,\ \cos^2(-x)=\cos^2x\Rightarrow f(-x)=-f(x)\). So \(\int_{-\pi}^{\pi} f(x)\,dx=0\).
161. 161. Evaluate \(I=\int_{-\pi}^{\pi}(x+x^3)\,dx\).
     Both \(x\) and \(x^3\) are odd ⇒ integral over symmetric interval is \(0\).
162. 162. \(f(x)=x^3\sin^6x\). Show odd ⇒ integral zero on \([-\pi/4,\pi/4]\).
     \(x^3\) odd, \(\sin^6x\) even ⇒ product odd ⇒ integral \(0\).
163. 163. \(I=\int_{-\pi/2}^{\pi/2}(x^3+x\cos x+2\tan^5x+3)\,dx\).
     Odd parts vanish; only constant term remains: \(\int_{-\pi/2}^{\pi/2}3\,dx=3\pi\).
164. 164. \(I=\int_{-5}^{5}(3x^2-x^{10}\sin x+x^5\sqrt{1+x^2})\,dx\).
     Split: \(\int_{-5}^5 3x^2dx=6\int_0^5 x^2dx=250\). The other two integrals are zero (integrands odd). So total \(I=250\).
165. 165. \(f(x)=(1-x^2)\sin x\cos^2x\). Show odd ⇒ integral zero.
     \(f(-x)=-(1-x^2)\sin x\cos^2x\) ⇒ odd ⇒ integral over symmetric interval is 0.
166. 166. \(I=\displaystyle\int_{-1}^{1}\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx\).
     Put \(x=\sin t\), \(dx=\cos t\,dt\). Then integrand becomes \(t\sin t\,dt\) with \(t\in[-\pi/2,\pi/2]\). \(I=2\int_0^{\pi/2} t\sin t\,dt\). Integration by parts gives \(\int_0^{\pi/2} t\sin t\,dt = 1/2\). So \(I=2\).
167. 167. \(I=\displaystyle\int_{-1}^{1}\frac{x^2\sin^{-1}x}{\sqrt{1-x^2}}\,dx\).
     \(x^2\) even, \(\sin^{-1}x\) odd ⇒ integrand odd ⇒ integral \(0\).
168. 168. \(\displaystyle\int_{-\pi/2}^{\pi/2}\sin^5x\,dx\).
     \(\sin^5x\) is odd ⇒ integral \(0\).
169. 169. \(\displaystyle\int_{-a}^{a} x\sqrt{a^2-x^2}\,dx\).
     Integrand is odd ⇒ integral \(0\).
170. 170. \(\displaystyle\int_{-\pi}^{\pi}\frac{x\cos x}{1+\sin^2x}\,dx\).
     Numerator odd, denominator even ⇒ integrand odd ⇒ integral \(0\).
171. 171. \(\displaystyle\int_{-1}^{1}\tan x\,dx\).
     \(\tan x\) is odd ⇒ integral \(0\) (interval avoids singularities).
172. 172. \(\displaystyle\int_{-\pi/2}^{\pi/2}\frac{dx}{\sin^3x+\sin x}\).
     Denominator odd ⇒ integrand odd ⇒ integral \(0\).
173. 173. \(\displaystyle\int_{-\pi/2}^{\pi/2}(3\sin x+\sin^3x)\,dx\).
     Both terms odd ⇒ integral \(0\).
174. 174. \(\displaystyle\int_{-1}^{1}(\sqrt{1+x+x^2}-\sqrt{1-x+x^2})\,dx\).
     The integrand is odd (substitute \(x\mapsto -x\)), hence the integral is \(0\).

Part 2: Q.175–194

175. 175. \(\displaystyle\int_{-\pi/2}^{\pi/2}\sin^{11}x\,dx\).
     \(\sin^{11}x\) is odd ⇒ integral \(0\).
176. 176. \(\displaystyle\int_{-1/2}^{1/2}\frac{x^2}{x^2-1}\,dx\).
     Integrand even ⇒ \(I=2\int_0^{1/2}\frac{x^2}{x^2-1}dx\). \(\frac{x^2}{x^2-1}=1+\frac{1}{x^2-1}\). \[ I=1+2\left[\tfrac{1}{2}\ln\left|\frac{x-1}{x+1}\right|\right]_0^{1/2} =1+\ln\left|\frac{-1/2}{3/2}\right|-0 =1-\ln 3. \]
177. 177. If \(F(x)=(f(x)+f(-x))(g(x)-g(-x))\), show integral is 0.
     \(F(-x)=-F(x)\) ⇒ odd ⇒ \(\int_{-\pi/2}^{\pi/2}F(x)\,dx=0\).
178. 178. \(I=\displaystyle\int_{-3\pi/2}^{-\pi/2}\big((x+\pi)^3+\cos^2(x+\pi)\big)\,dx\).
     Put \(t=x+\pi\). Limits: \(-\pi/2\) to \(\pi/2\). \[ I=\int_{-\pi/2}^{\pi/2}(t^3+\cos^2t)\,dt. \] \(\int_{-\pi/2}^{\pi/2}t^3dt=0\). And \(\int_{-\pi/2}^{\pi/2}\cos^2t\,dt=2\int_0^{\pi/2}\cos^2t\,dt =2\cdot\frac{\pi}{4}=\frac{\pi}{2}\). So \(I=\frac{\pi}{2}\).
179. 179. \(\displaystyle\int_{-\log 2}^{\log 2}\sin\!\Big(\frac{e^x-1}{e^x+1}\Big)\,dx\).
     Define \(f(x)=\sin\frac{e^x-1}{e^x+1}\). Then \(f(-x)=-f(x)\). Hence integral \(0\).
180. 180. \(\displaystyle\int_{-2}^{0}\big[(x+1)^3+2+(x+1)\cos(x+1)\big]\,dx\).
     Substitute \(t=x+1\). Limits \(-1\) to \(1\). Odd terms vanish; only constant \(2\) contributes: \(\int_{-1}^1 2\,dt = 4\).
181. 181. \(\displaystyle\int_{-\pi/4}^{\pi/4}\frac{e^x\sec^2x}{e^{2x}-1}\,dx\).
     Check \(f(-x)=-f(x)\) by direct substitution. So integrand odd ⇒ integral \(0\).
182. 182. \(\displaystyle\int_{-1}^{1}\log\frac{a-x}{a+x}\,dx\).
     \(f(-x)=-f(x)\) ⇒ integrand odd ⇒ integral \(0\).
183. 183. \(\displaystyle\int_{-1/2}^{1/2}\cos x\log\frac{1+x}{1-x}\,dx\).
     Under \(x\mapsto -x\) integrand changes sign ⇒ integral \(0\).
184. 184. \(\displaystyle\int_{-\pi/2}^{\pi/2}\log\frac{2-\sin x}{2+\sin x}\,dx\).
     Substitution \(x\mapsto -x\) shows integrand odd ⇒ integral \(0\).
185. 185. \(\displaystyle\int_{-a}^{a}\ln(x+\sqrt{1+x^2})\,dx\).
     The integrand is odd (see Q.157) ⇒ integral \(0\).
186. 186. \(\displaystyle\int_{-\pi/2}^{\pi/2}\sin\{\ln(x+\sqrt{1+x^2})\}\,dx\).
     Inside log is odd and \(\sin\) is odd ⇒ composite odd ⇒ integral \(0\).
187. 187. \(\displaystyle\int_{-\pi/2}^{\pi/2}\sin^2x\cos^2x(\sin x+\cos x)\,dx\).
     Split into two integrals. The \(\sin^3x\cos^2x\) part is odd (vanishes). The remaining part yields \[ I=2\int_0^{\pi/2}\sin^2x\cos^3x\,dx =2\Big(\int_0^{\pi/2}\cos^3x\,dx-\int_0^{\pi/2}\cos^5x\,dx\Big) =\frac{4}{15}. \]
188. 188. \(\displaystyle\int_{-a}^{a}\frac{\sin^2x}{1-x^2}\,dx\).
     Integrand even ⇒ \(2\int_0^a \frac{\sin^2x}{1-x^2}\,dx\).
189. 189. \(f(x)=\dfrac{x\sin x}{\cos^2x}\). Show even.
     \(f(-x)=\dfrac{x\sin x}{\cos^2x}=f(x)\) so even.
190. 190. \(\displaystyle\int_{-p}^{p}(a\tan^3x+b\cos^2x+c\sin x)\,dx\).
     Odd parts vanish. Only \(\cos^2x\) remains: \[ I=2b\int_0^p\cos^2x\,dx=b\Big(p+\frac{\sin2p}{2}\Big). \]
191. 191. \(\displaystyle\int_{-\pi/2}^{\pi/2}\frac{\sqrt{1-\cos2x}}{2}\,dx\).
     \(\sqrt{1-\cos2x}=\sqrt2\,|\sin x|\). So integral becomes \(\sqrt2\int_0^{\pi/2}\sin x\,dx=\sqrt2\).
192. 192. \(\displaystyle\int_{-1}^{1}\frac{d}{dx}\left(\tan^{-1}\frac{1}{x}\right)\,dx\).
     \(\frac{d}{dx}\tan^{-1}(1/x)=-\frac{1}{1+x^2}\). So integral equals \(-\int_{-1}^1\frac{1}{1+x^2}dx = -\frac{\pi}{2}\).
193. 193. \(\displaystyle\int_{-\pi}^{\pi}(\cos px-\sin qx)^2\,dx\).
     Expand and use orthogonality. Cross-term zero; each squared term integrates to \(\pi\). So result \(2\pi\).
194. 194. \(\displaystyle\int_0^{\alpha}\frac{dx}{(1+x^2)^2}\).
     Put \(x=\tan t\). Then integral becomes \(\int_0^{\arctan\alpha}\cos^2t\,dt\) \(=\left[\tfrac{t}{2}+\tfrac{\sin2t}{4}\right]_0^{\arctan\alpha}.\)

Part 3: Q.195–204

195. 195. \(\displaystyle\int_0^{\pi/2}\frac{dt}{1+2t\cos\alpha+t^2}\).
     This standard integral evaluates to \(\dfrac{\alpha}{2\sin\alpha}\) for \(0<\alpha<\pi\), derivable by tangent-half-angle or parameter differentiation.
196. 196. Show \(|\sin x|\) is even and \(\dfrac{\sin x}{1+\cos x}\) is odd.
     \(|\sin(-x)|=|\sin x|\) ⇒ even. And \(\dfrac{\sin(-x)}{1+\cos(-x)}=-\dfrac{\sin x}{1+\cos x}\) ⇒ odd.
197. 197. \(\displaystyle\int_0^{\pi/2}\frac{\cos x-\sin x}{1+\sin x\cos x}\,dx\).
     Under \(x\mapsto \tfrac{\pi}{2}-x\) numerator changes sign while denominator is invariant, so \(I=-I\Rightarrow I=0\).
198. 198. \(\displaystyle\int_0^{\pi/2}\sin2x\log\tan x\,dx\).
     Under \(x\mapsto \tfrac{\pi}{2}-x\), \(\log\tan x\) flips sign while \(\sin2x\) remains, giving \(I=-I\Rightarrow I=0\).
199. 199. \(\displaystyle\int_0^{\pi/2}\sin2x\log\cot x\,dx\).
     Same idea as Q.198 ⇒ integral \(0\).
200. 200. \(\displaystyle\int_0^{\pi}\frac{\sin2nx}{\sin x}\,dx\).
     Replace \(x\mapsto \pi-x\): numerator changes sign, denominator same ⇒ integrand negates ⇒ integral \(0\).
201. 201. \(\displaystyle\int_0^{\pi}\cos^3x\,dx\).
     Using \(x\mapsto\pi-x\), integrand negates ⇒ integral \(0\).
202. 202. \(\displaystyle\int_0^{\pi/2}\frac{\cos2x}{(\sin x+\cos x)^2}\,dx\).
     Under \(x\mapsto \tfrac{\pi}{2}-x\), integrand changes sign ⇒ integral \(0\).
203. 203. \(\displaystyle\int_0^{\pi} x\cos^2x\cos^3((2n+1)x)\,dx\).
     Use \(x\mapsto\pi-x\). The \(\cos^3((2n+1)x)\) changes sign (odd multiple), so integral cancels ⇒ \(0\).
204. 204. \(\displaystyle\int_0^{\pi} e^{\cos^2x}\cos^3((2n+1)x)\,dx\).
     \(e^{\cos^2x}\) is even under \(x\mapsto\pi-x\), while \(\cos^3((2n+1)x)\) changes sign ⇒ integrand odd ⇒ integral \(0\).