Question

integrate (3sinx-4cosx+5sec²x-2cosec²x) dx​

Answer 1

Formula Applied :

$\bullet\ \; \sf \int \sin x=-\cos x\\\\\bullet\ \; \sf \int \cos x=\sin x\\\\\bullet\ \; \sf \int \sec ^2x=\tan x\\\\\bullet\ \; \sf \int \csc^2 x=-\cot x$

Solution :

$\\ \displaystyle \sf \int \left(3.sinx-4.cosx+5.sec^2x-2.csc^2x\right)dx$

$\\ \to \displaystyle 3\int \sf sinx\ dx-4\int cosx\ dx+5\int sec^2x\ dx-2\int csc^2x\ dx$

$\\ \to \displaystyle \sf 3(-cosx)-4(sinx)+5(tanx)-2(-cotx) +c$

$\\ \to \sf -3cosx-4sinx+5tanx+2cotx+c$

$\\ \to \sf \pink{2cotx-3cosx-4sinx+5tanx+c}\ \; \bigstar$

More Info :

Integration and differentiation are both opposite to each other .

Let's see some examples ,

$\bullet\ \; \sf \dfrac{d}{dx}(sin\ x)=cos\ x\\\\\bullet\ \; \sf \int cos\ x=sin\ x$

$\bullet\ \; \sf \dfrac{d}{dx}(cos\ x)=-sin\ x \\\\\bullet\ \; \sf \int sin\ x=-cos\ x$

Answer 2

Explanation:

Formula Applied :

\begin{gathered}\bullet\ \; \sf \int \sin x=-\cos x\\\\\bullet\ \; \sf \int \cos x=\sin x\\\\\bullet\ \; \sf \int \sec ^2x=\tan x\\\\\bullet\ \; \sf \int \csc^2 x=-\cot x\end{gathered}

∙ ∫sinx=−cosx

∙ ∫cosx=sinx

∙ ∫sec

2

x=tanx

∙ ∫csc

2

x=−cotx

Solution :

\begin{gathered}\\ \displaystyle \sf \int \left(3.sinx-4.cosx+5.sec^2x-2.csc^2x\right)dx\\\end{gathered}

∫(3.sinx−4.cosx+5.sec

2

x−2.csc

2

x)dx

\begin{gathered}\\ \to \displaystyle 3\int \sf sinx\ dx-4\int cosx\ dx+5\int sec^2x\ dx-2\int csc^2x\ dx \\\end{gathered}

→3∫sinx dx−4∫cosx dx+5∫sec

2

x dx−2∫csc

2

x dx

\begin{gathered}\\ \to \displaystyle \sf 3(-cosx)-4(sinx)+5(tanx)-2(-cotx) +c \\\end{gathered}

→3(−cosx)−4(sinx)+5(tanx)−2(−cotx)+c

\begin{gathered}\\ \to \sf -3cosx-4sinx+5tanx+2cotx+c\\\end{gathered}

→−3cosx−4sinx+5tanx+2cotx+c

\begin{gathered}\\ \to \sf \pink{2cotx-3cosx-4sinx+5tanx+c}\ \; \bigstar \\\end{gathered}

→2cotx−3cosx−4sinx+5tanx+c ★

More Info :

Integration and differentiation are both opposite to each other .

Let's see some examples ,

\begin{gathered}\bullet\ \; \sf \dfrac{d}{dx}(sin\ x)=cos\ x\\\\\bullet\ \; \sf \int cos\ x=sin\ x\end{gathered}

∙

dx

d

(sin x)=cos x

∙ ∫cos x=sin x

\begin{gathered}\bullet\ \; \sf \dfrac{d}{dx}(cos\ x)=-sin\ x \\\\\bullet\ \; \sf \int sin\ x=-cos\ x\end{gathered}

∙

dx

d

(cos x)=−sin x

∙ ∫sin x=−cos x