ত্রিকোনোমিতিক ফাংশনের যোগজীকরণ
∫1−cos5x1+cos5x=?\int \frac{1-\cos 5 x}{1+\cos 5 x}= ?∫1+cos5x1−cos5x=?
−25tan5x2−x+c-\frac{2}{5} \tan \frac{5 x}{2}-x+c−52tan25x−x+c
25tan5x2−x+c\frac{2}{5} \tan \frac{5 x}{2}-x+c52tan25x−x+c
25tanx2−x+c\frac{2}{5} \tan \frac{ x}{2}-x+c52tan2x−x+c
25tan5x3+x+c\frac{2}{5} \tan \frac{5 x}{3}+x+c52tan35x+x+c
Solve:
∫1−cos5x1+cos5x=∫2sin25x22cos25x2dx=∫tan25x2dx=∫(sec25x2−1)dx=25tan5x2−x+c \begin{array}{l} \int \frac{1-\cos 5 x}{1+\cos 5 x} \\ =\int \frac{2 \sin ^{2} \frac{5 x}{2}}{2 \cos ^{2} \frac{5 x}{2}} d x=\int \tan ^{2} \frac{5 x}{2} d x \\ =\int\left(\sec ^{2} \frac{5 x}{2}-1\right) d x=\frac{2}{5} \tan \frac{5 x}{2}-x+c \end{array} ∫1+cos5x1−cos5x=∫2cos225x2sin225xdx=∫tan225xdx=∫(sec225x−1)dx=52tan25x−x+c
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
