Integration of (3sinx-2)cosx/13-cos^2x-7sinx
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Integration of (3sinx-2)cosx/13-cos^2x-7sinx
Please find this answer I= \[\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 - \cos^2 x - 7\sin x}dx\] = \[\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 -(1 - \ sin ^2 x) - 7\sin x}dx\] `( cos^2x =1 - sin^2 x)`\[ = \int\frac{\left( 3\sin x - 2 \right) \cos x}{\text{ sin^}2 x - 7\sin x + 12}dx\]\[ = \int\frac{\left( 3\sin x - 2 \right) \cos x}{\text{ sin^}2 x - 4\sin x 3\text{ sin } x + 12}dx\]\[ = \int\frac{\left( 3\sin x - 2 \right) \cos x}{\sin x\left( \sin x - 4 \right) - 3\left( \sin x - 4 \right)}dx\]\[ = \int\frac{\left( 3\sin x - 2 \right)\cos x}{\left( \sin x - 3 \right)\left( \sin x - 4 \right)}dx\] \[\text{ Let sin x }= t\]\[ \Rightarrow \text{ cos x dx }= dt\]\[ \therefore I = \int\frac{\left( 3t - 2 \right)}{\left( t - 3 \right)\left( t - 4 \right)}dt\] Using partial fraction, we get \[\frac{\left( 3t - 2 \right)}{\left( t - 3 \right)\left( t - 4 \right)} = \frac{A}{\left( t - 3 \right)} + \frac{B}{\left( t - 4 \right)} = \frac{A\left( t - 4 \right) + B\left( t - 3 \right)}{\left( t - 3 \right)\left( t - 4 \right)}\]\[ \Rightarrow 3t - 2 = (A + B)t - 4A - 3B\] Comparing coefficients, we get A = - 7 and B = 10 So, \[I = 7\int\frac{1}{\left( t - 3 \right)}dt + 10\int\frac{1}{\left( t - 4 \right)}dt\] \[\Rightarrow I = - 7\text{ ln }\left| t - 3 \right| + 10\text{ ln}\left| t - 4 \right| + c\]\[ \therefore I = - 7\text{ ln }\left| \sin x - 3 \right| + 10 \text{ ln }\left| \sin x - 4 \right| + c\]Page 2 \[\int\frac{x + 7}{3 x^2 + 25x + 28}\text{ dx}\] \[I = \int\frac{x + 7}{3 x^2 + 25x + 28}dx\]\[ = \int\frac{x + 7}{3 x^2 + 21x + 4x + 28}dx\]\[ = \int\frac{x + 7}{3x\left( x + 7 \right) + 4\left( x + 7 \right)}dx\]\[ = \int\frac{x + 7}{\left( 3x + 4 \right)\left( x + 7 \right)}dx\] \[= \int\frac{1}{(3x + 4)}dx\]\[ = \frac{1}{3}\text{ ln }\left| 3x + 4 \right| + c\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 3\ [I = \int\frac{x^3}{x^4 + x^2 + 1}dx\]\[ = \int\frac{x^2 \cdot x}{\left( x^2 \right)^2 + x^2 + 1}dx\]\[\text{ Let x }^2 = \text{ t or 2xdx } = dt\]\[ \Rightarrow I = \frac{1}{2}\int\frac{t}{t^2 + t + 1}dt\]\[ = \frac{1}{4}\int\frac{2t}{t^2 + t + 1}dt\]\[ = \frac{1}{4}\int\frac{2t + 1 - 1}{t^2 + t + 1}dt\] \[= \frac{1}{4}\int\left[ \frac{\left( 2t + 1 \right)}{\left( t^2 + t + 1 \right)} - \frac{1}{\left( t^2 + t + 1 \right)} \right]dt\]\[ = \frac{1}{4}\left[ \text{ log}\left| t^2 + t + 1 \right| - \int\frac{1}{\left( t^2 + t + \frac{1}{4} + \frac{3}{4} \right)}dt \right]\]\[ = \frac{1}{4}\left[ \text{ log }\left| t^2 + t + 1 \right| - \int\frac{1}{\left( t + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}dt \right]\]\[ = \frac{1}{4}\left[ \text{ log}\left| t^2 + t + 1 \right| - \frac{2}{\sqrt{3}}\tan\frac{\left( t + \frac{1}{2} \right)}{\left( \frac{\sqrt{3}}{2} \right)} \right] + c\]\[ = \frac{1}{4}\left[ \text{ log }\left| t^2 + t + 1 \right| - \frac{2}{\sqrt{3}}\tan\left( \frac{2t + 1}{\sqrt{3}} \right) \right] + c\] \[= \frac{1}{4}\left[ \text{ log }\left| x^4 + x^2 + 1 \right| - \frac{2}{\sqrt{3}}\tan\left( \frac{2 x^2 + 1}{\sqrt{3}} \right) \right] + c\] Q. 44.0( 4 Votes )Class 12thRD Sharma - Volume 119. Indefinite IntegralsAnswer :I = = I = Let, sin x = t cos x dx = dt I = As we can see that there is a term of t in numerator and derivative of t2 is also 2t. So there is a chance that we can make substitution for t2 ? 7t + 12 and I can be reduced to a fundamental integration.As, Let, 3t ? 2 = A(2t ? 7) + B 3t ? 2 = 2At ? 7A + BOn comparing both sides ?We have,2A = 3 A = 3/2?7A + B = ?2 B = 7A ? 2 = 17/2Hence,I = I = Let, I1 = and I2 = Now, I = I1 + I2 ....eqn 1We will solve I1 and I2 individually.As, I1 = Let u = t2 ? 7t + 12 du = (2t ? 7)dx I1 reduces to Hence,I1 = { }On substituting value of u, we have:I1 = ....eqn 2As, I2 = and we don't have any derivative of function present in denominator. we will use some special integrals to solve the problem.As denominator doesn't have any square root term. So one of the following two integrals will solve the problem. I2 = I2 = Using: a2 ? 2ab + b2 = (a ? b)2We have:I2 =I2 matches with the form I2 =I2 =...eqn 3From eqn 1, we have:I = I1 + I2Using eqn 2 and 3, we get ?I = Putting value of t in I:I = .....ansRate this question :How useful is this solution?We strive to provide quality solutions. Please rate us to serve you better.PREVIOUSEvaluate the inteNEXTEvaluate the inteTry our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic ExpertsDedicated counsellor for each studentDetailed Performance Evaluationview all courses The questions posted on the site are solely user generated, Doubtnut has no ownership or control over the nature and content of those questions. Doubtnut is not responsible for any discrepancies concerning the duplicity of content over those questions.
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