Calculus Solution Ex#5.3

by

 

 

EXERCISE 5 \cdot 3

Evaluate the integrals using appropriate Substitutions

15-

    \[\begin{aligned} \int(4 x-3)^{9} d x \\ \\ \text { Let } u=4 x-3 \quad \quad \quad (1) \\ \\ d u=4 d x \\ \\ \frac{1}{4} d u=d x \end{aligned}\]

then the above integral becomes

    \[\begin{aligned} \frac{1}{4} \int u^{9} d u & =\frac{1}{4}\left[\frac{u^{9+1}}{9+1}\right]+c \\ \\ & =\frac{1}{40} u^{10}+c-(2) \\ \\ \end{aligned}\]

From (1) substituting u=4 x-3 into (2).

    \[\int(4 x-3)^{9} d x=\frac{1}{40}(4 x-3)^{10}+c\]

16-

    \[\begin{aligned} \int x^{3} \sqrt{(5+x^{4})} d x \\ \\ \int x^{3}\left(5+x^{4}\right)^{1 / 2} d x-(1) \\ \\ \text { Let } u=5+x^{4}----(2) \\ \\ d u=4 x^{3} d x \\ \\ \frac{1}{4} du=x^{3} dx \end{aligned}\]

then the integral (1) becomes

    \[\begin{aligned} \frac{1}{4} \int u^{1 / 2} d u & =\frac{1}{4}\left[\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]+c \\ \\ & =\frac{1}{4}\left[\frac{u^{3 / 2}}{3 / 2}\right]+c \\ \\ & =\frac{1}{6} u^{3 / 2}+c ---(3) \end{aligned}\]

Now substituting u=5+x^{4} into (3)

    \[\int x^{3} \sqrt{5+x^{4}} d x=\frac{1}{6}\left[5+x^{4}\right]^{3 / 2}+c\]

17-

    \[\begin{aligned} \int \sin 7 x d x \\ \\ \text {Let} \quad u=7 x \\ \\ d u=7 d x \\ \\ \frac{1}{7} d u=d x \end{aligned}\]

Then the above integral becomes

    \[\begin{aligned} \frac{1}{7} \int \sin u d u & =\frac{1}{7}(-\cos u)+c \\ \\ & =-\frac{1}{7} \cos u+c -----(1) \end{aligned}\]

Substituting u=7 x into (1)

    \[\int \sin 7 x d x=-\frac{1}{7} \cos 7 x+c\]

18-

    \[\begin{aligned} \int \cos \frac{x}{3} d x \\ \\ \text {Let} \quad u=\frac{x}{3} \\ \\ d u=\frac{1}{3} d x \Rightarrow 3 d u=d x \end{aligned}\]

Then the above integral becomes

    \[\end{aligned} 3 \int \cos u d u=3 \sin u+c ----(1) \end{aligned}\]

Substituting u=x / 3 into (1)

    \[\int \cos \frac{x}{3} d x=3 \sin \frac{x}{3}+c\]

19-

    \[\begin{aligned} \int \sec 4 x \tan 4 x d x \\ \\ \tex {Let} \quad u=4 x \\ \\ d u= 4 dx \Rightarrow \frac{1}{4} d u=dx \end{aligned}\]

Then the above integral becomes

    \[\frac{1}{4} \int \sec u \tan u d u=\frac{1}{4} \sec u+c ---(1)\]

Substituting u=4 x into (1)

    \[\int \sec 4 x \tan 4 x d x=\frac{1}{4} \sec 4 x+c\]

20-

    \[\begin{aligned} \int \sec ^{2} 5 x d x$ \\ \\ \text {Let} \quad u=5 x \Rightarrow d u=5 d x \Rightarrow \frac{1}{5} d u=d x \end{aligned}\]

Then the above integral becomes

    \[\frac{1}{5} \int \sec ^{2} u d u=\frac{1}{5} \tan u+c ----(1)\]

Substituting u=5 x into (1)

    \[\int \sec ^{2} 5 x d x=\frac{1}{5} \tan 5 x+c\]

23-

    \[\begin{aligned} \int \frac{d x}{\sqrt{1-4 x^{2}}} \\ \\ \text {Let} \quad u=2 x \\ \\ d u=2 d x \Rightarrow \frac{1}{2} d u=d x \end{aligned}\]

Then the above integral becomes

    \[\begin{aligned} \int \frac{\frac{1}{2} d u}{\sqrt{1-u^{2}}} & =\frac{1}{2} \int \frac{d u}{\sqrt{1-u^{2}}} \\ \\ & =\frac{1}{2} \sin ^{-1}(u)+c ----(1) \end{aligned}\]

Substituting u=2 x into (1)

    \[\int \frac{d x}{\sqrt{1-4 x^{2}}}=\frac{1}{2} \sin ^{-1}(2 x)+c\]

28-

    \[\begin{aligned} \int \frac{x^{2}+1}{\sqrt{x^{3}+3 x}} d x \\ \\ \text {Let} \quad u=x^{3}+3 x \\ \\ d u & =(3 x^{2}+3) d x \Rightarrow \frac{1}{3} d u=(x^{2}+1) d x \\ \\ \text {Then the above integral becomes} \\ \\ \frac{1}{3} \int \frac{d u}{\sqrt{u}}=\frac{1}{3} \int \frac{d u}{u^{1 / 2}} \\ \\ =\frac{1}{3} \int u^{-1 / 2} d u \\ \\ =\frac{1}{3} \left[\frac{u^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} \right]+c \\ \\ =\frac{1}{3}\left[\frac{u^{\frac{1}{2}}}{\frac{1}{2}}\right]+c \\ \\ =\frac{2}{3} u^{\frac{1}{2}}+c ------(1) \end{aligned}\]

Substituting u=x^{3}+3 x into (1)

    \[\int \frac{x^{3}+3 x}{\sqrt{x^{3}+3 x}} d x=\frac{2}{3}\left(x^{3}+3 x\right)^{\frac{1}{2}}+c\]