Tekhnic Integration by part 1) Find ∫x.sec^2 x dx 2) Find ∫x.e^(2x) dx 3) Find ∫ln x dx 4) Find ∫x^2 .e^(2x) dx 5) Find ∫e^x cos x dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 156008 by zainaltanjung last updated on 07/Oct/21

$Tekhnic Integration by part$ $1) Find \int x . \sec^{2} x dx$ $2) Find \int x . e^{2 x} dx$ $3) Find \int \ln x dx$ $4) Find \int x^{2} . e^{2 x} dx$ $5) Find \int e^{x} \cos x dx$
Commented by SANOGO last updated on 07/Oct/21

$c o o l$
Commented by puissant last updated on 07/Oct/21

$1)$ $A = \int x {s e c}^{2} x d x = x t a n x - \int t a n x$ $= x t a n x - (- l n ∣ c o s x ∣) + C$ $\Rightarrow A = x t a n x + l n ∣ c o s x ∣ + C . .$
Commented by puissant last updated on 07/Oct/21
$2) S=∫x e^(2x) dx = (1/2)x e^(2x) −(1/2)∫e^(2x) dx =(1/2)x e^(2x) −(1/4)e^(2x) +C ⇒ S= (1/2){x e^(2x) −(1/2)e^(2x) }+C$
$2)$ $S = \int x e^{2 x} d x = \frac{1}{2} x e^{2 x} - \frac{1}{2} \int e^{2 x} d x$ $= \frac{1}{2} x e^{2 x} - \frac{1}{4} e^{2 x} + C$ $\Rightarrow S = \frac{1}{2} {x e^{2 x} - \frac{1}{2} e^{2 x}} + C$
Commented by SANOGO last updated on 07/Oct/21

$3) \int l n x = x l n x - x + c$
Commented by puissant last updated on 07/Oct/21

$3)$ $I = \int l n x d x$ ${\begin{cases} u = l n x \\ v^{'} = 1 \end{cases} \Rightarrow {\begin{cases} u^{'} = \frac{1}{x} \\ v = x \end{cases}$ $\Rightarrow I = x l n x - \int 1 d x$ $\Rightarrow I = x l n x - x + C$
Commented by puissant last updated on 07/Oct/21
$4) J=∫x^2 e^(2x) dx { ((u=x^2 )),((v′=e^(2x) )) :} ⇒ { ((u′=2x)),((v=(1/2)e^(2x) )) :} ⇒ J = (x^2 /2)e^(2x) −∫xe^(2x) dx ⇒ J=(x^2 /2)e^(2x) −(x/2)e^(2x) +(1/4)e^(2x) +C$
$4)$ $J = \int x^{2} e^{2 x} d x$ ${\begin{cases} u = x^{2} \\ v^{'} = e^{2 x} \end{cases} \Rightarrow {\begin{cases} u^{'} = 2 x \\ v = \frac{1}{2} e^{2 x} \end{cases}$ $\Rightarrow J = \frac{x^{2}}{2} e^{2 x} - \int {x e}^{2 x} d x$ $\Rightarrow J = \frac{x^{2}}{2} e^{2 x} - \frac{x}{2} e^{2 x} + \frac{1}{4} e^{2 x} + C$
Commented by puissant last updated on 07/Oct/21
$5) D = ∫e^x cosx dx { ((u=cosx)),((v′=e^x )) :} ⇒ { ((u′=−sinx)),((v=e^x )) :} ⇒ D = e^x cosx+∫e^x sinx dx { ((u=sinx)),((v′=e^x )) :} ⇒ { ((u′=cosx)),((v=e^x )) :} ⇒ D = e^x cosx+e^x sinx−∫e^x cosxdx ⇒ D = e^x (cosx+sinx)−D ⇒ 2D=e^x (cosx+sinx) ⇒ D=(e^x /2)(cosx+sinx)+C..$
$5)$ $D = \int e^{x} c o s x d x$ ${\begin{cases} u = c o s x \\ v^{'} = e^{x} \end{cases} \Rightarrow {\begin{cases} u^{'} = - s i n x \\ v = e^{x} \end{cases}$ $\Rightarrow D = e^{x} c o s x + \int e^{x} s i n x d x$ ${\begin{cases} u = s i n x \\ v^{'} = e^{x} \end{cases} \Rightarrow {\begin{cases} u^{'} = c o s x \\ v = e^{x} \end{cases}$ $\Rightarrow D = e^{x} c o s x + e^{x} s i n x - \int e^{x} c o s x d x$ $\Rightarrow D = e^{x} (c o s x + s i n x) - D$ $\Rightarrow 2 D = e^{x} (c o s x + s i n x)$ $\Rightarrow D = \frac{e^{x}}{2} (c o s x + s i n x) + C . .$
Commented by tabata last updated on 08/Oct/21

$3) w i t h o u t b y p a r t$ $\int l n x d x = \int (l n x + 1 - 1) d x$ $= \int ((l n x + 1) - 1) d x$ $= \int d (x l n x) - \int d x$ $= x l n x - x + C$ $◻ M$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum