Antiderivative Of Secx Antiderivative Of Secx Hello friends today we will discuss how to find the antiderivative of sec x for finding the antiderivative of secx we will use some identities of trigonometry,substitution method and the partial fraction method the antiderivative of secx is also known as as integration of secx let me show you what is the actual mean of partial fraction and the way of its use. For finding antiderivative secx use the trigonometric identity secx = 1/cosx This implies that secx = cosx/1-sin2x It means ∫ secx dx = ∫ cosx/1-sin2x dx Now use the substitution method of integration for sinx Let sinx =t by differentiating it we get cosx dx = dt Put the value of sinx and dx in above integral equation so that we get Know More About Cooling Formula
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∫ cosx/1-sin2x dx = ∫ t/(1-t2) dt Now we use the partial fraction method Let t/(1-t2) = A/(1-t) + B/(1+t) by solving it we will get A + B =1 and At – Bt =0 By solving these equation for A and B we will get A = ½ and B = ½ By substituting the value of A and B in the equation t/(1-t2) = A/(1-t) + B/(1+t) We will get t/(1-t2) = 1/2(1-t) + 1/2(1+t) By doing integration both side we get ∫t/(1-t2) = ∫1/2(1-t) + ∫1/2(1+t) It implies that ∫t/(1-t2) = -1/2 Log(1-t) +1/2 Log(1+t) Log x
because ∫1/x =
It implies that ∫t/(1-t2) =Log((1+t)/(1-t))1/2 since Log x – Log y = Log(x/y) By substituting the value of t in the equation we will get ∫ secx dx = ∫ cosx/1-sin2x dx = Log(secx+ tanx) so finally we get the result that antiderivative secx = Log(secx+ tanx) as we seen the antiderivative of secx become quite easier by the method we used known as partial fraction.by other method may be the antiderivative secx is quite typical and confusing. Read More About Appropriate Measure Of Central Tendency
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se the Previous Year Question Papers For Functional Englishof Cbse Board. For checking the result is correct or not, we differentiate the result with respect to x. Check- let the result y = Log(secx+ tanx) Here we use some derivative of a function such that derivative of logx = 1/x, derivative of secx is ( secx *tanx ) and derivative of tanx = sec2x and we will also use the method of function of a function. Differentiate y = Log(secx+ tanx) with respect to x We get dy/dx = 1/(secx+ tanx) ) * d/dx ( secx +tanx ) since d/dx Logx = 1/x On further differentiating by using the method function of function We get dy/dx = 1/ (secx + tanx ) * ( secx*tanx + sec2x ) * d/dx (x) since derivative of the function secx = secx + tanx and derivative of the function tanx = sec2x By simplifying we get dy/dx = ( secx*tanx + sec2x ) / (secx + tanx ) We get secx (tanx + secx) / (secx + tanx ) As a result we get secx As we can say that derivative of Log(secx+ tanx) is secx We know that derivative and antiderivative of a function are converse of each other .Therefore we can say that the antiderivative secx = Log(secx+ tanx) Since secx = 1/cosx and tanx = sinx /cosx Log(secx+ tanx) can be written as Log( 1/cosx + sinx/cosx ) It implies that Log (1+sinx)/cosx So we can say that antiderivative secx = Log(secx+ tanx) = Log (1+sinx)/cosx Tutorcircle.com
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