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\(\int cos^3x \,sinx\,dx = - \int t^3 dt\)   \(\begin{pmatrix}\because \text{Let} \, cos x = t\\⇒-sinx\,dx = dt\\⇒sinx\,dx = - dt\end{pmatrix}\) 

\(= \frac{-t^4}4 + C\)

\(= \frac{-cos^4x}{4}+C\)      \((\because t = cos x)\) 

\(\int (10w^4 + 9w^3 + 7w)dw\)

\(= 10 \int w^4dw + 9\int w^3 dw +7\int wdw\)

\(= 10\left(\frac{w^5}{5}\right)+ 9\left(\frac{w^4}{4}\right)+ 7\left(\frac{w^2}2\right)+ C\)

\(= 2w^5 + \frac94w^4 + \frac72w^2 + C\)

\(\int (6x^5 -18x^2 + 7)dx\)

\(= 6\int x^5 dx - 18\int x^2 dx + 7\int dx\)

\(= 6\left(\frac{x^6}{6}\right)-18\left(\frac{x^3}3\right) + 7x + C\)

\(= x^6 - 6x^3 + 7x + C\)

\(\int\frac{dx}{\sqrt{8x+3}+\sqrt{8x+1}}\)

 = \(\int\frac{\sqrt{8x+3}-\sqrt{8x+1}}{(8x+3)-(8x+1)}dx\) 

 = \(\int\frac{\sqrt{8x+3}-\sqrt{8x+1}}{2}dx\) 

 = \(\frac12[\int\sqrt{8x+3}dx-\int\sqrt{8x+1}dx]\) 

 = \(\frac12[\frac23\frac{(8x+3)^{3/2}}8-\frac23\frac{(8x+1)^{3/2}}8]+c\)

 = \(\frac12\times\frac23\times\frac18[(8x+3)^{3/2}-(8x+1)^{3/2}]+c\) 

 = \(\frac1{24}[(8x+3)^{3/2}-(8x+1)^{3/2}]+c\)

Assume,

sinx = t 

d(sinx) = dt 

cosx dx = dt 

∴ Substituting t and dt in given equation we get

⇒ \(\int\)t⁵dt

⇒ \(\frac{t^6}{6}\) + c

But,

t = sinx

⇒ \(\frac{sin^6x}{6}\) + c

Formula to be used - \(\int\cfrac{dx}{\sqrt{x^2\pm a^2}}=log(x+\sqrt{x^2\pm a^2})+c\), where c is the integrating constant

\(\therefore\int\cfrac{dx}{\sqrt{4x^2-1}}\)

\(= \int\cfrac{dx}{\sqrt{(2x)^{2}-1^2}}\)

\(=\cfrac{1}{2}log|2x+\sqrt{4x^2-1}|+c\), c being the integrating constant

\(\frac{1+tan\,x}{1-tan\,x}\) = \(\frac{cos\,x+sin\,x}{cos\,x-sin\,x}\)

\(\int\frac{cos\,x+sin\,x}{cos\,x-sin\,x}dx\)

Let cos x - sin x = t 

-(sin x + cos x)dx = dt

\(\int\frac{-dt}{t}\) = - int + c

\(\int\frac{1+tan\,x}{1-tan\,x}dx\) = - In[cosx - sinx] + c

Formula to be used -\(\int\cfrac{dx}{\sqrt{x^2\pm a^2}}=log(x+\sqrt{x^2\pm a^2})+c\)

\(\therefore\int\cfrac{dx}{\sqrt{9x^2-7}}\)

\(=\int\cfrac{dx}{\sqrt{(3x)^{2}-\sqrt{7}^2}}\)

\(=log|3x+\sqrt{9x^2-7}|+c\), c being the integrating constant

Correct answer is (a) \(\frac{x}{1 + log\,x} + c\)

1/2∫ 2sin3x sin x dx

{ 2 sin A sin B=cos(A-B)-cos(A+B) }

1/2∫ (cos2x - cos4x) dx = sin2x/4 - sin4x/8 + c

∫xe^3x+5 dx

= x.\(\frac{e^{3x + 5}}{3} - \int\frac{e^{3x + 5}}{3}.1dx\)

= \(\frac{xe^{3x + 5}}{3} - \frac{e^{3x + 5}}{9} + c\)

We can write as ∫3log₃x² dx

By simplifying, we get

∫x² dx

On integrating, we get

x³/3 + c

Given as ∫log_x x dx

Given equation, we can write as

∫1. dx

On integrating, we get

x + c