ত্রিকোনোমিতিক ফাংশনের যোগজীকরণ
∫tanx+cotxdx=? \int \sqrt{\tan{x}} + \sqrt{\cot{x}} dx = ? ∫tanx+cotxdx=?
∫(tanx+cotx)dx=∫(sinxcosx+cosxsinx)dx=∫(sinx+cosxsinx⋅cosx)dx=∫2(sinx+cosx)dx2sinxcosx=∫2(sinx+cosx)1−(1−2sinxcosx)dx=∫2(sinx+cosx)1−(sin2x+cos2x−2sinxcosx)=∫2(sinx+cosx)dx1−(sinx−cosx)2∣ ধরি, −cosx+sinx=z⇒dzdx=(sinx+cosx)⇒(sinx+cosx)dx=dz তাহলে, ∫2dz1−z2=2sin−1z+c=2sin−1(sinx−cosx)+c (Ans.) \begin{array}{l} \int(\sqrt{\tan x}+\sqrt{\cot x}) d x=\int\left(\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}}\right) d x=\int\left(\frac{\sin x+\cos x}{\sqrt{\sin x \cdot \cos x}}\right) d x \\ =\int \frac{\sqrt{2}(\sin x+\cos x) d x}{\sqrt{2 \sin x \cos x}}=\int \frac{\sqrt{2}(\sin x+\cos x)}{\sqrt{1-(1-2 \sin x \cos x)}} d x=\int \frac{\sqrt{2}(\sin x+\cos x)}{\sqrt{1-\left(\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x\right)}} \\ =\int \frac{\sqrt{2}(\sin x+\cos x) d x}{\sqrt{1-(\sin x-\cos x)^{2}}} \mid \text { ধরি, }-\cos x+\sin x=z \Rightarrow \frac{d z}{d x}=(\sin x+\cos x) \Rightarrow(\sin x+\cos x) d x=d z \\ \text { তাহলে, } \int \frac{\sqrt{2} d z}{\sqrt{1-z^{2}}}=\sqrt{2} \sin ^{-1} z+c=\sqrt{2} \sin ^{-1}(\sin x-\cos x)+c \text { (Ans.) }\end{array} ∫(tanx+cotx)dx=∫(cosxsinx+sinxcosx)dx=∫(sinx⋅cosxsinx+cosx)dx=∫2sinxcosx2(sinx+cosx)dx=∫1−(1−2sinxcosx)2(sinx+cosx)dx=∫1−(sin2x+cos2x−2sinxcosx)2(sinx+cosx)=∫1−(sinx−cosx)22(sinx+cosx)dx∣ ধরি, −cosx+sinx=z⇒dxdz=(sinx+cosx)⇒(sinx+cosx)dx=dz তাহলে, ∫1−z22dz=2sin−1z+c=2sin−1(sinx−cosx)+c (Ans.)
f(x)=x………(i) f(x)=x \ldots \ldots \ldots(i) f(x)=x………(i)
g(x)=cos−1x2………(ii) g(x)=\cos ^{-1} x^2 \ldots \ldots \ldots(i i) g(x)=cos−1x2………(ii)
y2=7x………(iii) y^2=7 x \ldots \ldots \ldots(i i i) y2=7x………(iii)
∫dx1+cosx=f(x)+c \int \frac{dx}{1 + \cos{x}} = f{\left ( x \right )} + c ∫1+cosxdx=f(x)+c হলে, f(x)=?
∫cos−1xdx \int \cos^{- 1}{x} dx ∫cos−1xdx এর মান কোনটি?
f(x)=xsin−1x এবং g(x)=16−x2 \mathrm{f(x)=x \sin ^{-1} x \text { এবং } g(x)=\sqrt{16-x^{2}}} f(x)=xsin−1x এবং g(x)=16−x2
