নির্দিষ্ট যোগজ
∫0π/4(tan3x+tanx)dx\int_{0}^{\pi / 4}\left(\tan ^{3} x+\tan x\right) d x∫0π/4(tan3x+tanx)dx
12\frac{1}{2}21
−12-\frac{1}{2}−21
14\frac{1}{4}41
16\frac{1}{6}61
Solve:
∫0π/4(tan3x+tanx)dx=∫0π/4(tan2x+1)tanxdx \begin{array}{l} \int_{0}^{\pi / 4}\left(\tan ^{3} x+\tan x\right) d x \\ =\int_{0}^{\pi / 4}\left(\tan ^{2} x+1\right) \tan x d x \end{array} ∫0π/4(tan3x+tanx)dx=∫0π/4(tan2x+1)tanxdx
=∫0π/4sec2xtanxdx=∫0π/4(tanx)d(tanx)=[12(tanx)2]0π/4=12{(tanπ4)2−(tan0)2}=12{(1)2−0}=12 \begin{array}{l} =\int_{0}^{\pi / 4} \sec ^{2} x \tan x d x \\ =\int_{0}^{\pi / 4}(\tan x) d(\tan x)=\left[\frac{1}{2}(\tan x)^{2}\right]_{0}^{\pi / 4} \\ =\frac{1}{2}\left\{\left(\tan \frac{\pi}{4}\right)^{2}-(\tan 0)^{2}\right\}=\frac{1}{2}\left\{(1)^{2}-0\right\}=\frac{1}{2} \end{array} =∫0π/4sec2xtanxdx=∫0π/4(tanx)d(tanx)=[21(tanx)2]0π/4=21{(tan4π)2−(tan0)2}=21{(1)2−0}=21
The value of ∫−π/2199π/2(1+cos2x)dx\displaystyle\int^{199\pi/2}_{-\pi/2}\sqrt{(1+\cos 2x)}dx∫−π/2199π/2(1+cos2x)dx is?
∫0π/2cosxdx= কত? \int_{0}^{\pi / 2} \cos x d x=\text { কত? } ∫0π/2cosxdx= কত?
∫0π6sin2xcosxdx= \int_{0}^{\frac{\pi}{6}} \sin ^{2} x \cos x d x= ∫06πsin2xcosxdx= কত ?
∫1/21dxx4x2−1=?\int_{1 / 2}^{1} \frac{d x}{x \sqrt{4 x^{2}-1}}=?∫1/21x4x2−1dx=?
