দৃশ্যকল্প-১ হতে নির্ণয় কর : (i) $\int_0^{\pi / 2} \sqrt{1+\mathrm{g}( heta)} \mathrm{d} heta$; $ \int_0^{\pi / 2} f( heta) \sqrt[3]{g( heta)} d heta $

Question

দৃশ্যকল্প-১ হতে নির্ণয় কর : (i) $\int_0^{\pi / 2} \sqrt{1+\mathrm{g}(	heta)} \mathrm{d} 	heta$; $ \int_0^{\pi / 2} f(	heta) \sqrt[3]{g(	heta)} d 	heta $

Accepted Answer

(i) $\begin{aligned} & \int_0^{\pi / 2} \sqrt{1+\mathrm{g}(	heta)} \mathrm{d} 	heta \ &= \int_0^{\pi / 2} \sqrt{1+\sin 	heta} \mathrm{d} 	heta \ &= \int_0^{\pi / 2} \sqrt{\sin^2 \frac{	heta}{2}+\cos^2 \frac{	heta}{2}+2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}} \mathrm{~d} 	heta \ &= \int_0^{\pi / 2} \sqrt{\left(\sin \frac{	heta}{2}+\cos \frac{	heta}{2}ight)^2} \mathrm{~d} 	heta \ &=\int_0^{\pi / 2}\left(\sin \frac{	heta}{2}+\cos \frac{	heta}{2}ight) d 	heta \ &=\left.-\frac{\cos \frac{	heta}{2}}{\frac{1}{2}}+\frac{\sin \frac{	heta}{2}}{\frac{1}{2}}ight]_0^{\frac{\pi}{2}} \ &=-2 \cos \frac{\pi}{4}+2 \sin \frac{\pi}{4}+2 \cos 0-2 \sin 0 \ &=-2 \cdot \frac{1}{\sqrt{2}}+2 \cdot \frac{1}{\sqrt{2}}+2-0 \ &= 2(	ext{ Ans. })\end{aligned}$ ধরি, $ \mathrm{z}=\sin 	heta $ $ 	herefore \mathrm{dz}=\cos 	heta \mathrm{d} 	heta $ $\begin{aligned} & 	ext{ (ii) } \int_0^{\pi / 2} f(	heta) \sqrt[3]{g(	heta)} d 	heta \ & =\int_0^{\pi / 2} \cos^3 	heta \sqrt[3]{\sin 	heta} d 	heta \ & =\int_0^{\pi / 2}\left(1-\sin^2 	hetaight) \sqrt[3]{\sin 	heta} \cdot \cos 	heta d 	heta \ & =\int_0^1\left(1-z^2ight) z^{\frac{1}{3}} d z \ & =\int_0^1\left(z^{\frac{1}{3}}-z^{\frac{7}{3}}ight) d z\end{aligned}$ $\begin{aligned} & =\left[\frac{z^{\frac{1}{3}+1}}{\frac{1}{3}+1}-\frac{z^{\frac{7}{3}}+1}{\frac{7}{3}+1}ight]_0^1 \ & =\left[\frac{3}{4} \cdot z^{\frac{4}{3}}-\frac{3}{10} z^{\frac{10}{3}}ight]_0^1 \ & =\frac{3}{4}-\frac{3}{10}-0+0 \ & =\frac{15-6}{20}=\frac{9}{20} 	ext{ (Ans.) }\end{aligned}$

HSC Higher Math 1st Paper CQ — দৃশ্যকল্প-১ হতে নির্ণয় কর : (i) $\int_0^{\pi / 2} \sqrt{1+\mathrm{g}(\t

আরো প্রশ্ন দেখো