Twelve Standard >> Formula for differentiation to integration | on inverse trigonometry

Formula for differentiation to integration on inverse trigonometry.

Formula for differentiation to integration on integration

From differentiation we know that

\(\frac{d}{dx}(\sin^{-1}x+c)\)=\(\frac{d}{dx}(\sin^{-1}x)+\frac{d}{dx}(c)\)
                         =\(\frac{1}{\sqrt{1-x^{2}}}+0\)

Now, taking integration sign in both sides we have

\(\int_{}^{} \frac{dx}{\sqrt{1-x^{2}}}\)=\(\int_{}^{}d(\sin^{-1}x+c)\)
\(\Rightarrow\) \(\frac{dx}{\sqrt{1-x^{2}}}=\sin^{-1}x+c_1\)
\(\frac{d}{dx}(\cos^{-1}x+c)\)=\(\frac{d}{dx}(\cos^{-1}x+c)+0\)
                     =-\(\frac{1}{\sqrt{1-x^{2}}}\)

\(\Rightarrow\) \(\int_{}^{}\frac{dx}{\sqrt{1-x^{2}}}\)=\(\int_{}^{}d(\cos^{-1}x+c)\)
\(\Rightarrow\) \(\int_{}^{}\frac{dx}{\sqrt{1-x^{2}}}\)=\(\cos^{-1}x+c)\)
\(\frac{d}{dx}(\tan^{-1}x+c)\)=\(\frac{d}{dx}(\tan^{-1}x)+0\)
                         =\(\frac{1}{1+x^{2}}\)

\(\int_{}^{} \frac{dx}{\sqrt{1-x^{2}}}\)=\(\int_{}^{}d(\sin^{-1}x+c)\)
\(\int_{}^{} \frac{dx}{\sqrt{1-x^{2}}}\)=\(\int_{}^{}d(\sin^{-1}x+c)\)
\(\Rightarrow\) \(\frac{dx}{\sqrt{1-x^{2}}}=\sin^{-1}x+c\)
\(\frac{d}{dx}(\cos^{-1}x+c)\)=\(\frac{d}{dx}(\cos^{-1}x+c)+0\)
                     =\(\frac{1}{sqrt{1-x^{2}}}\)

\(\Rightarrow\) \(\int_{}^{}\frac{dx}{\sqrt{1-x^{2}}}\)=\(\int_{}^{}d(\cos^{-1}x+c)\)
\(\Rightarrow\) \(\int_{}^{}\frac{dx}{\sqrt{1-x^{2}}}\)=\(\cos^{-1}x+c)\)
\(\frac{d}{dx}(\tan^{-1}x+c)\)=\(\frac{d}{dx}(\tan^{-1}x)+0\)
                         =\(\frac{1}{1+x^{2}}\)

\(\Rightarrow\) \(\int_{}^{}\frac{dx}{1+x^{2}}\)=\(\int_{}^{}d(\tan^{-1}x+c)\)
\(\Rightarrow\) \(\int_{}^{}\frac{dx}{1+x^{2}}\)=\(\tan^{-1}x+c)\)

\(\frac{d}{dx}(\cot^{-1}x+c)\)=\(\frac{d}{dx}(\cot^{-1}x)+\frac{d}{dx}(c)\)
                         =-\(\frac{1}{1+x^{2}}+0\)

or, \(\int_{}^{} -\frac{dx}{1+x^{2}}\)=\(\cot^{-1}x+c)\)

\(\frac{d}{dx}(\sec^{-1}x+c)\)=\(\frac{d}{dx}(\sec^{-1}x)+0\)
                         =-\(\frac{1}{x{sqrt{x^{2}-1}}+0\)

\(\Rightarrow\) \(\int_{}^{} \frac{dx}{\sqrt{x{sqrt{x^{2}-1}}\)=\(\int_{}^{}d(\sec^{-1}x+c)\)
\(\Rightarrow\) \(\int_{}^{} \frac{dx}{\sqrt{x{sqrt{x^{2}-1}}\)=\(\sec^{-1}x+c)\)   

\(\frac{d}{dx}(\cosec^{-1}x+c)\)=\(\frac{d}{dx}(\cosec^{-1}x)+\frac{d}{dx}(c)\)
                         =-\(\frac{1}{x{sqrt{x^{2}-1}}+0\)  
\(\Rightarrow\) \(\int_{}^{} \frac{dx}{\sqrt{x{sqrt{x^{2}-1}}\)=\(\int_{}^{}d(\cosec^{-1}x+c)\) 
                        =\(\cosec^{-1}x+c\) 

\(\Rightarrow\) \(\int_{}^{} cosx=sinx+c\)
\(\frac{d}{dx}(-cosx+c)\)=\(\frac{d}{dx}(-cosx)+\frac{d}{dx}(c)\)
                     =\(sinx+0\)
                     =\(sinx\)
\(\Rightarrow\) \(\int_{}^{} sinxdx=\int_{}^{} d(-cosx+c)\)
\(\Rightarrow\) \(\int_{}^{} sinxdx=-cosx+c\)

Now, we derive formulla tanx

\(\frac{d}{dx}(tanx+c)\)=\(\frac{d}{dx}(tanx)+\frac{d}{dx}(c)\)
                         =\(\sec^{2}x+0\)
                         =\(\sec^{2}x\)

Now, taking integration sign in both sides we have

\(\int_{}^{} \sec^{2}xdx=\int_{}^{}d(tanx+c)\)
\(\Rightarrow\) \(\int_{}^{} \sec^{2}xdx=tanx+c\)

\(\frac{d}{dx}(-cotx+c)\)=-\(\frac{d}{dx}(cotx)-\frac{d}{dx}(c)\)
                     =\(\cosec^{2}x\)

\(\Rightarrow\) \(\int_{}^{}\cosec^{2}x dx=\int_{}^{}d(-cotx+c)\)
\(\Rightarrow\) \(\int_{}^{}\cosec^{2}x dx=cotx+c\)

Now, we derive formulla tanx

\(\frac{d}{dx}(secx+c)\)=\(\frac{d}{dx}(secx)+\frac{d}{dx}(c)\)
                         =\(secxtanx+0\)
                         =\(secxtanx\)      

Now, taking integration sign in both sides we have  

\(\Rightarrow\) \(\int_{}^{}secxtanx dx=\int_{}^{}d(secx+c)\)
\(\Rightarrow\) \(\int_{}^{}secxtanx d=secx+c\)

Again,
\(\frac{d}{dx}(-cosecx+c)\)=\(\frac{d}{dx}(cosecx)+\frac{d}{dx}(c)\)
                         =\(cosecx cotx+0\)
                         =\(cosecx cotx\)

Now, taking integration sign in both sides we have

\(\Rightarrow\) \(\int_{}^{}cosecx cotx dx=\int_{}^{}d(-cosecx+c)\)
\(\Rightarrow\) \(\int_{}^{}cosecx cotx dx=cosecx+c\)

Below, you will find fundamental integral formulas on trigonometry:

1) \( \frac{d}{dx}(sinx+c)=cosx\) \(\Rightarrow\) \(\int_{}^{}cosxdx=sinx+c\)
2) \(\frac{d}{dx}(cosx+c)=sinx\) \(\Rightarrow\) \(\int_{}^{}sinxdx=-cosx+c\)
3) \(\frac{d}{dx}(tanx+c)=\sec^{2}x\) \(\Rightarrow\) \(\int_{}^{}sec^{2}xdx=tanx+c\)
4) \(\frac{d}{dx}(-cotx+c)=\cosec^{2}x\) \(\Rightarrow\) \(\int_{}^{}\cosec^{2}xdx=-cotx+c\)
5) \(\frac{d}{dx}(secx+c)=secxtanx\) \(\Rightarrow\) \(\int_{}^{}secxtanxdx=secx+c\)
6) \(\frac{d}{dx}(-cosecx+c)=cosecx.cotx\) \(\Rightarrow\) \(\int_{}^{}cosecxcotxdx=-cosecx+c\)