বিভিন্ন সূত্রের ব্যবহারে যোগজীকরণ
∫sinpxcosqxdx,(p>q)=?\int \sin p x \cos q x d x,(p>q) = ?∫sinpxcosqxdx,(p>q)=?
−12{sin(p+q)xp+sin(p−q)xp}+c-\frac{1}{2}\left\{\frac{\sin (p+q) x}{p}+\frac{\sin (p-q) x}{p}\right\}+c−21{psin(p+q)x+psin(p−q)x}+c
−12{cos(p+q)xp+cos(p−q)xp}+c-\frac{1}{2}\left\{\frac{\cos (p+q) x}{p}+\frac{\cos (p-q) x}{p}\right\}+c−21{pcos(p+q)x+pcos(p−q)x}+c
−12{sin(p+q)xp+q+sin(p−q)xp−q}+c-\frac{1}{2}\left\{\frac{\sin (p+q) x}{p+q}+\frac{\sin (p-q) x}{p-q}\right\}+c−21{p+qsin(p+q)x+p−qsin(p−q)x}+c
−12{cos(p+q)xp+q+cos(p−q)xp−q}+c-\frac{1}{2}\left\{\frac{\cos (p+q) x}{p+q}+\frac{\cos (p-q) x}{p-q}\right\}+c−21{p+qcos(p+q)x+p−qcos(p−q)x}+c
Solve:
∫sinpxcosqxdx,(p>q)=∫12{sin(p+q)x+sin(p−q)x}dx=12{−cos(p+q)xp+q−cos(p−q)xp⋯q}+c=−12{cos(p+q)xp+q+cos(p−q)xp−q}+c \begin{array}{l} \int \sin p x \cos q x d x,(p>q) \\ =\int \frac{1}{2}\{\sin (p+q) x+\sin (p-q) x\} d x \\ =\frac{1}{2}\left\{-\frac{\cos (p+q) x}{p+q}-\frac{\cos (p-q) x}{p \cdots q}\right\}+c \\ =-\frac{1}{2}\left\{\frac{\cos (p+q) x}{p+q}+\frac{\cos (p-q) x}{p-q}\right\}+c \end{array} ∫sinpxcosqxdx,(p>q)=∫21{sin(p+q)x+sin(p−q)x}dx=21{−p+qcos(p+q)x−p⋯qcos(p−q)x}+c=−21{p+qcos(p+q)x+p−qcos(p−q)x}+c
∫sin(5−x10)dx=f(x)+c \int \sin{\left ( 5 - \frac{x}{10} \right )} dx = f{\left ( x \right )} + c ∫sin(5−10x)dx=f(x)+c হলে, f(x)এর মান কত?
∫dx25−x2=? \int \frac{d x}{\sqrt{25-x^{2}}} \quad = ? ∫25−x2dx=?
g(x)=x g(x)=\sqrt{x} g(x)=x হলে-
i. ∫1g(x)dx=2x+c \int \frac{1}{g(x)} d x=2 \sqrt{x}+c ∫g(x)1dx=2x+c
ii. ∫01g(x)dx=23 \int_{0}^{1} g(x) d x=\frac{2}{3} ∫01g(x)dx=32
iii. ∫sec2xdxg(tanx)=2tanx+c \int \frac{\sec ^{2} x d x}{g(\tan x)}=2 \sqrt{\tan x}+c ∫g(tanx)sec2xdx=2tanx+c
নিচের কোনটি সঠিক?
∫x9x4+4dx=? \int \frac{x}{9 x^{4} + 4} dx = ? ∫9x4+4xdx=?
