In Exercises 1-4, evaluate the integral by following the...

In Exercises $1-4,$ evaluate the integral by following the steps given. $$I=\int \frac{d x}{x^{2} \sqrt{x^{2}-2}}$$ (a) Show that the substitution $x=\sqrt{2} \sec \theta$ transforms the integral $I$ into $\frac{1}{2} \int \cos \theta d \theta,$ and evaluate $I$ in terms of $\theta$ . (b) Use a right triangle to show that with the above substitution, $\sin \theta=\sqrt{x^{2}-2} / x$ (c) Evaluate $I$ in terms of $x$ .

In Exercises $1-4,$ evaluate the integral by following the steps given.
$$I=\int \frac{d x}{x^{2} \sqrt{x^{2}-2}}$$
(a) Show that the substitution $x=\sqrt{2} \sec \theta$ transforms the integral $I$ into $\frac{1}{2} \int \cos \theta d \theta,$ and evaluate $I$ in terms of $\theta$ .
(b) Use a right triangle to show that with the above substitution, $\sin \theta=\sqrt{x^{2}-2} / x$
(c) Evaluate $I$ in terms of $x$ .

Key Concepts

Right Triangle Relationships

Right triangle relationships connect the sides and angles of a right triangle, allowing one to express trigonometric functions in terms of ratios of sides. This is particularly useful in trigonometric substitution, where constructing a right triangle based on the substitution helps to translate back to the original variable by linking the trigonometric functions to algebraic expressions.

Inverse Trigonometric Functions

Inverse trigonometric functions arise when reversing the trigonometric substitution process. They are used to express the antiderivative in terms of the original variable by taking the inverse of the trigonometric function, thus converting the solution obtained in the substitution variable back to the initial context.

Integration with Substitution

Integration with substitution is a technique where one changes the variable of integration to simplify the integrand. This often involves replacing a complex expression with a new variable, computing its differential, and thereby transforming the integral into a more manageable form.

Trigonometric Substitution

Trigonometric substitution is a method used to simplify integrals containing algebraic expressions, especially those involving square roots. By replacing a variable with a trigonometric function (such as replacing x with a multiple of sec ?, sine, or tangent), the radical expression is transformed using trigonometric identities, resulting in a form that is easier to integrate.

Transcript

00:02 Integrate i that is integration of 1 over x square root of x square minus 2 d x now if we see the function is in the denominator so we have to simplify before integrating such function and to simplify we can first look at into the trigonometric functions and for this we know if we substitute x equals to root 2 sec theta we might get a solution.

00:36 Okay.

00:38 So since we have x well as well as dx, so we have to differentiate x with respect to theta.

00:44 So differentiate x with respect to theta.

00:48 We'll have dx over d theta is equal to root two is constant and differentiation of sec theta is sec theta times 10 theta great so we have both the values okay so we need d x so we can shift d theta on the right side so we'll have d x is root 2 sec theta 10 theta d theta so let's substitute this value here x square so that is root 2 sec theta square root of x square it is again root 2 sec theta square minus 2 and value of d x is root 2 sec theta tan theta d theta okay so now we have to simplify this it looks a bit more dirtier than the question given but it will get simplified in the next step let's see how so integration 1 over square of root 2 is 2 square of sec theta is sec squared theta and inside the root again square of root 2 is 2 square theta is sec square theta minus 2 and here we have root 2 sec set theta tan theta great now we'll multiply this 1 with this numerator term so we have root 2 sec theta tan theta and we'll keep this d theta in multiplication here and in denominator we have 2 sex squared theta if we take 2 out it will come out as root 2 and we'll have 6 squared theta minus 1 from trigonometry we know 1 plus 10 square theta is 6 square theta this is identity.

03:15 So sex square theta minus 1 will be 10 square theta.

03:25 So this will become root 2 sec theta, 10 theta over 2, sex square theta, root 2, 6 squared theta minus 1 is 10 square theta.

03:41 So this becomes 10 square theta d theta.

03:46 Now we can do some cancellations here.

03:49 We find that root 2 is in numerator and denomero.

03:51 Denominator we can cancel one of the sec theta and square and root get cancelled and this is tan theta get cancelled with tan theta so we are left with one over two sec theta d theta that's a great simplification if you see so we can take sec theta in numerator so we have a property that is cost theta is 1 over sec theta okay this is 1 by 2 cost theta d theta great so we have simplified it now we can easily integrate we can take half out as it is a constant value and then integrate cos theta so we know integration of cost theta is minus sine theta sorry it's sine theta and we'll add a constant c that represent it is indefinite integration.

05:15 Okay, so that is our answer for i.

05:19 But if you look, the answer is in terms of theta and question provided was in terms of x.

05:26 So we'll go back to our substitution...

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