Question
Evaluate the following integrals:$\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-\cos^2\text{x}-4\sin\text{x}}\text{ dx}$
✓
Answer
Let $\text{I}=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-\cos^2\text{x}-4\sin\text{x}}\text{ dx}$ $\therefore\ \text{I}=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-(1-\sin^2\text{x})-4\sin\text{x}}\text{ dx}$ $\Rightarrow\text{I}=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-1+\sin^2\text{x}-4\sin\text{x}}\text{ dx}$ Substitute $\sin\text{x}=\text{t}$ $\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$ Thus, $\text{I}=\int\frac{(3\text{t}-2)}{4+\text{t}^2-4\text{t}}\text{ dt}$ $\text{I}=\int\frac{(3\text{t}-2)}{\text{t}^2-4\text{t}+4}\text{ dt}$ $\text{I}=\int\frac{(3\text{t}-2)}{(\text{t}-2)^2}\text{ dt}$ Now let us separate the integrand into the simplest form using partial fractions. $\frac{(3\text{t}-2)}{(\text{t}-2)^2}=\frac{\text{A}}{(\text{t}-2)}+\frac{\text{B}}{(\text{t}-2)^2}$ $=\frac{\text{A}(\text{t}-2)+\text{B}}{(\text{t}-2)^2}$ $=\frac{\text{At}-2\text{A}+\text{B}}{(\text{t}-2)^2}$ $\Rightarrow3\text{t}-2=\text{At}-2\text{A}+\text{B}$ Comparing the coefficients, we have, $\text{A}=3$and
$-2\text{A}+\text{B}=-2$ Substituting the value of A = 3 in the above equation, we have, $\Rightarrow-2\times3+\text{B}=-2$ $\Rightarrow-6+\text{B}=-2$ $\Rightarrow\text{B}=6-2$ $\Rightarrow\text{B}=4$Thus, $\text{I}=\int\frac{(3\text{t}-2)}{(\text{t}-2)^2}\text{ dt}$ becomes,
$\text{I}=\int\frac{3}{(\text{t}-2)^2}\text{ dt}+\int\frac{4}{(\text{t}-2)^2}\text{ dt}$ $=3\log|\text{t}-2|-4\Big(\frac{1}{\text{t}-2}\Big)+\text{C}$ $=3\log|2-\text{t}|+4\Big(\frac{1}{2-\text{t}}\Big)+\text{C}$ Now, substituting $\text{t}=\sin\text{x},$ we have, $=3\log|2-\sin\text{x}|+4\Big(\frac{1}{2-\sin\text{x}}\Big)+\text{C}$
$-2\text{A}+\text{B}=-2$ Substituting the value of A = 3 in the above equation, we have, $\Rightarrow-2\times3+\text{B}=-2$ $\Rightarrow-6+\text{B}=-2$ $\Rightarrow\text{B}=6-2$ $\Rightarrow\text{B}=4$Thus, $\text{I}=\int\frac{(3\text{t}-2)}{(\text{t}-2)^2}\text{ dt}$ becomes,
$\text{I}=\int\frac{3}{(\text{t}-2)^2}\text{ dt}+\int\frac{4}{(\text{t}-2)^2}\text{ dt}$ $=3\log|\text{t}-2|-4\Big(\frac{1}{\text{t}-2}\Big)+\text{C}$ $=3\log|2-\text{t}|+4\Big(\frac{1}{2-\text{t}}\Big)+\text{C}$ Now, substituting $\text{t}=\sin\text{x},$ we have, $=3\log|2-\sin\text{x}|+4\Big(\frac{1}{2-\sin\text{x}}\Big)+\text{C}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Using properties of determinants, prove that
; background:rgba(220,220,220,0.5))
Differentiate the following functions with respect to x:
$\text{x}^{\cos^{-1}\text{x}}$
→$\text{x}^{\cos^{-1}\text{x}}$
If four points A, B, C and D with position vectors $4\hat{\text{i}}+3\hat{\text{j}},5\hat{\text{i}}+\text{x}\hat{\text{j}}+7\hat{\text{k}},5\hat{\text{i}}+3\hat{\text{j}}$ and $7\hat{\text{i}}+6\hat{\text{j}}+\hat{\text{k}}$ respectively are coplanar, then find the value of x.
→The resultant of two forces $\vec{\text{P}}$ and $\vec{\text{Q}}$ acting at an angle $\theta$is equal to $(2m+1)\sqrt{P^{2}+Q^{2}}$ and when they act at a angle $\Bigg(\frac{\pi}{2}-\theta\Bigg)$the resultant is equal to $(2m-1)\sqrt{P^{2}+Q^{2}}$. Show that $\tan\theta=\frac{m-1}{m+1}.$
→Find the value of x if $\begin{bmatrix}1&\text{x}&1\end{bmatrix}\begin{bmatrix}1&3&2\\2&5&1\\15&3&2\end{bmatrix}\begin{bmatrix}1\\2\\\text{x}\end{bmatrix}=0$
→Evaluate the following:
$\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\begin{pmatrix}\begin{bmatrix}1&0&2\\2&0&1\end{bmatrix}-\begin{bmatrix}0&1&2\\1&0&2 \end{bmatrix}\end{pmatrix}$
→$\begin{bmatrix}1&-1\\0&2\\2&3\end{bmatrix}\begin{pmatrix}\begin{bmatrix}1&0&2\\2&0&1\end{bmatrix}-\begin{bmatrix}0&1&2\\1&0&2 \end{bmatrix}\end{pmatrix}$
Let $A =\{1,2,3\}$ and $R =\left\{( a , b ): a , b \in A\right.$ and $\left|a^2-b^2\right| \leq 5$. Write R as set of ordered pairs. Mention whether $R$ is i. reflexive
$ii.$ symmetric
$iii.$ transitive
Give reason in each case.
→$ii.$ symmetric
$iii.$ transitive
Give reason in each case.
Find the points on the curve $y = x^3$ where the slope of the tangent is equal to the $x-$coordinate of the point.
→Arun and Tarun appeared for an interview for two vacancies. The probability of Arun's selection is $\frac{1}{4}$ and that to Tarun's rejection is $\frac{2}{3}$. Find the probability that at least one of them will be selected.
→Find the angles which the vector $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\sqrt{2}\hat{\text{k}}$ makes with the coordinate axes.
→